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Class 11 Mathematics Case Study Questions

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If you’re seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations. Furthermore, it is routinely updated to bring you up to speed with the newest CBSE syllabus. So, why delay? Begin your path to success with myCBSEguide now!

The rationale behind teaching Mathematics

The general rationale to teach Mathematics at the senior secondary level is to assist students:

• In knowledge acquisition and cognitive understanding of basic ideas, words, principles, symbols, and mastery of underlying processes and abilities, notably through motivation and visualization.
• To experience the flow of arguments while demonstrating a point or addressing an issue.
• To use the information and skills gained to address issues using several methods wherever possible.
• To cultivate a good mentality in order to think, evaluate, and explain coherently.
• To spark interest in the subject by taking part in relevant tournaments.
• To familiarise pupils with many areas of mathematics utilized in daily life.
• To pique students’ interest in studying mathematics as a discipline.

Case studies in Class 11 Mathematics

A case study in mathematics is a comprehensive examination of a specific mathematical topic or scenario. Case studies are frequently used to investigate the link between theory and practise, as well as the connections between different fields of mathematics. A case study will frequently focus on a specific topic or circumstance and will investigate it using a range of methodologies. These approaches may incorporate algebraic, geometric, and/or statistical analysis.

Sample Class 11 Mathematics case study questions

When it comes to preparing for Class 11 Mathematics, one of the best things Class 11 Mathematics students can do is to look at some Class 11 Mathematics sample case study questions. Class 11 Mathematics sample case study questions will give you a good idea of the types of Class 11 Mathematics sample case study questions that will be asked in the exam and help you to prepare more effectively.

Looking at sample questions is also a good way to identify any areas of weakness in your knowledge. If you find that you struggle with a particular topic, you can then focus your revision on that area.

myCBSEguide offers ample Class 11 Mathematics case study questions, so there is no excuse. With a little bit of preparation, Class 11 Mathematics students can boost their chances of getting the grade they deserve.

Some samples of Class 11 Mathematics case study questions are as follows:

Class 11 Mathematics case study question 1

• 9 km and 13 km
• 9.8 km and 13.8 km
• 9.5 km and 13.5 km
• 10 km and 14 km
• x  ≤   −1913
• x <  −1613
• −1613  < x <  −1913
• There are no solution.
• y  ≤   12 x+2
• y >  12 x+2
• y  ≥   12 x+2
• y <  12 x+2

Answer Key:

• (b) 9.8 km and 13.8 km
• (a) −1913   ≤  x 
• (b)  y >  12 x+2
• (d) (-5, 5)

Class 11 Mathematics case study question 2

• 2 C 1 × 13 C 10
• 2 C 1 × 10 C 13
• 1 C 2 × 13 C 10
• 2 C 10 × 13 C 10
• 6 C 2​ × 3 C 4   × 11 C 5 ​
• 6 C 2​ × 3 C 4   × 11 C 5
• 6 C 2​ × 3 C 5 × 11 C 4 ​
• 6 C 2 ​  ×   3 C 1 ​  × 11 C 5 ​
• (b) (13) 4  ways
• (c) 2860 ways.

Class 11 Mathematics case study question 3

Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the rate of Rs 1000 /yard². The builder planned ground floor of 5 m height, first floor of 4.75 m and so on each floor is 0.25 m less than its previous floor.

Class 11 Mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

• 5x + 6y = 60
• 6x + 5y = 60
• (a) (10, 0)
• (b) 6x + 5y = 60
• (b) 60/√ 61 km
• (d) 2√61 km

A peek at the Class 11 Mathematics curriculum

The Mathematics Syllabus has evolved over time in response to the subject’s expansion and developing societal requirements. The Senior Secondary stage serves as a springboard for students to pursue higher academic education in Mathematics or professional subjects such as Engineering, Physical and Biological Science, Commerce, or Computer Applications. The current updated curriculum has been prepared in compliance with the National Curriculum Framework 2005 and the instructions provided by the Focus Group on Teaching Mathematics 2005 in order to satisfy the rising demands of all student groups. Greater focus has been placed on the application of various principles by motivating the themes from real-life events and other subject areas.

Class 11 Mathematics (Code No. 041)

Design of Class 11 Mathematics exam paper

CBSE Class 11 mathematics question paper is designed to assess students’ understanding of the subject’s essential concepts. Class 11 mathematics question paper will assess their problem-solving and analytical abilities. Before beginning their test preparations, students in Class 11 maths should properly review the question paper format. This will assist Class 11 mathematics students in better understanding the paper and achieving optimum scores. Refer to the Class 11 Mathematics question paper design provided.

 Class 11 Mathematics Question Paper Design

• No chapter-wise weightage. Care to be taken to cover all the chapters.
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.  

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections.

  Prescribed Books:

• Mathematics Textbook for Class XI, NCERT Publications
• Mathematics Exemplar Problem for Class XI, Published by NCERT
• Mathematics Lab Manual class XI, published by NCERT

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CBSE Case Study Questions for Class 11 Maths Sets Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 11 Maths Sets in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 11 Maths Sets PDF

Checkout our case study questions for other chapters.

• Chapter 2 Relations and Functions Case Study Questions
• Chapter 3 Trigonometric Functions Case Study Questions
• Chapter 4 Principle of Mathematical Induction Case Study Questions
• Chapter 5 Complex Numbers and Quadratic Equations Case Study Questions

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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Chapter 3 Class 11 Trigonometric Functions

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NCERT Solutions of Chapter 3 Class 11 Trigonometry is available free at teachoo. You can check the detailed explanation of all questions of exercises, examples and miscellaneous by clicking on the Exercise link below.

We had learned Basics of Trigonometry in Class 10. In this chapter, we will learn

• What is a positive or a negative angle
• Measuring angles in Degree , Minutes and Seconds
• Radian measure of an angle
• Converting Degree to Radians , and vice-versa
• Sign of sin, cos, tan in all 4 quadrants
• Finding values of trigonometric functions when one value is given (Example: Finding value of sin, cot, cosec, tan, sec, when cos x = -3/5 is given)
• Finding Value of trigonometric functions, given angle
• Solving questions by formula like  (x + y) formula, 2x 3x formula, Cos x + cos y formula , 2 sin x sin y formula 
• Finding principal and general solutions of a trigonometric equation
• Sin and Cosine Formula with supplementary Questions

Important questions are marked, and Formula sheet is also provided. Click on an exercise or topic to begin.

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CBSE Class 11 Maths – Chapter 3 Trigonometric Functions- Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Sets : Notes and Study Materials -pdf

• Concepts of  Trigonometric Functions
• Trigonometric Functions Master File
• Trigonometric Functions Revision Notes
• R D Sharma Solution of Trigonometric Functions
• NCERT Solution  Trigonometric Functions
• NCERT  Exemplar Solution Trigonometric Functions
• Trigonometric Functions : Solved Example 1

CBSE Class 11 Maths Notes Chapter 3 Trigonometric Functions

Angle Angle is a measure of rotation of a given ray about its initial point. The original ray is called the initial side and the final position of ray after rotation is called terminal side of the angle. The point of rotation is called vertex. If the direction of rotation is anti-clockwise, the angle is said to be positive and if the direction of rotation is clockwise, then the angle is negative.

Measuring Angles There are two systems of measuring angles Sexagesimal system (degree measure): If a rotation from the initial side to terminal side is  ( 1 360 ) t h  of a revolution, the angle is said to have a measure of one degree, written as 1°. One sixtieth of a degree is called a minute, written as 1′ and one-sixtieth of a minute is called a second, written as 1″ Thus, 1° = 60′ and 1′ = 60″

Circular system (radian measure):  A radian is an angle subtended at the centre of a circle by an arc, whose length is equal to the radius of the circle. We denote 1 radian by 1°.

Relation Between Radian and Degree We know that a complete circle subtends at its centre an angle whose measure is 2π radians as well as 360°. 2π radian = 360°. Hence, π radian = 180° or 1 radian = 57° 16′ 21″ (approx) 1 degree = 0.01746 radian

Six Fundamental Trigonometric Identities

• sinx =  1 c o s e c x
• cos x =  1 s e c x
• tan x =  1 c o t x
• sin 2  x + cos 2  x = 1
• 1 + tan 2 x = sec 2  x
• 1 + cot 2  x = cosec 2  x

Trigonometric Functions – Class 11 Maths Notes

Trigonometric ratios are defined for acute angles as the ratio of the sides of a right angled triangle. The extension of trigonometric ratios to any angle in terms of radian measure (real number) are called trigonometric function. The signs of trigonometric function in different quadrants have been given in following table.

Domain and Range of Trigonometric Functions

Sine, Cosine, and Tangent of Some Angles Less Than 90°

Allied or Related Angles The angles  n π 2 ± θ  are called allied or related angle and θ ± n × (2π) are called coterminal angles. For general reduction, we have following rules, the value of trigonometric function for ( n π 2 ± θ ) is numerically equal to

• the value of the same function, if n is an even integer with the algebraic sign of the function as per the quadrant in which angle lies.
• the corresponding co-function of θ, if n is an odd integer with the algebraic sign of the function for the quadrant in which it lies, here sine and cosine, tan and cot, sec and cosec are cofunctions of each other.

Functions of Negative Angles

For any acute angle of θ. We have,

• sin(-θ) = – sinθ
• cos (-θ) = cosθ
• tan (-θ) = – tanθ
• cot (-θ) = – cotθ
• sec (-θ) = secθ
• cosec (-θ) = – cosecθ

Some Formulae Regarding Compound Angles

An angle made up of the sum or difference of two or more angles is called compound angles. The basic results in direction are called trigonometric identities as given below: (i) sin (x + y) = sin x cos y + cos x sin y (ii) sin (x – y) = sin x cos y – cos x sin y (iii) cos (x + y) = cos x cos y – sin x sin y (iv) cos (x – y) = cos x cos y + sin x sin y

(ix) sin(x + y) sin (x – y) = sin 2  x – sin 2  y = cos 2  y – cos 2  x (x) cos (x + y) cos (x – y) = cos 2  x – sin 2  y = cos 2  y – sin 2  x

Transformation Formulae

• 2 sin x cos y = sin (x + y) + sin (x – y)
• 2 cos x sin y = sin (x + y) – sin (x – y)
• 2 cos x cos y = cos (x + y) + cos (x – y)
• 2 sin x sin y = cos (x – y) – cos (x + y)
• sin x + sin y = 2 sin( x + y 2 ) cos( x − y 2 )
• sin x – sin y = 2 cos( x + y 2 ) sin( x − y 2 )
• cos x + cos y = 2 cos( x + y 2 ) cos( x − y 2 )
• cos x – cos y = -2 sin( x + y 2 ) sin( x − y 2 )

Trigonometric Ratios of Multiple Angles

Product of Trigonometric Ratios

• sin x sin (60° – x) sin (60° + x) =  1 4  sin 3x
• cos x cos (60° – x) cos (60° + x) =  1 4  cos 3x
• tan x tan (60° – x) tan (60° + x) = tan 3x
• cos 36° cos 72° =  1 4
• cos x . cos 2x . cos 2 2 x . cos 2 3 x … cos 2 n-1  =  s i n 2 n x 2 n s i n x

Sum of Trigonometric Ratio, if Angles are in A.P.

Trigonometric Equations Equation which involves trigonometric functions of unknown angles is known as the trigonometric equation.

Solution of a Trigonometric Equation A solution of a trigonometric equation is the value of the unknown angle that satisfies the equation. A trigonometric equation may have an infinite number of solutions.

Principal Solution The solutions of a trigonometric equation for which 0 ≤ x ≤ 2π are called principal solutions.

General Solutions A solution of a trigonometric equation, involving ‘n’ which gives all solution of a trigonometric equation is called the general solutions.

General Solutions of Trigonometric Equation

• sin x = 0 ⇔ x = nπ, n ∈ Z
• cos x = 0 ⇔ x = (2n + 1)  π 2  , n ∈ Z
• tan x = 0 ⇔ x = nπ, n ∈ Z
• sin x = sin y ⇔ x = nπ + (-1) n  y, n ∈ Z
• cos x = cos y ⇔ x = 2nπ ± y, n ∈ Z
• tan x = tan y ⇔ x = nπ ± y, n ∈ Z
• sin 2  x = sin 2  y ⇔ x = nπ ± y, n ∈ Z
• cos 2  x = cos 2  y ⇔ x = nπ ± y, n ∈ Z
• tan 2  x = tan 2  y ⇔ x = nπ ± y, n ∈ Z

Basic Rules of Triangle

In a triangle ABC, the angles are denoted by capital letters A, B and C and the lengths of sides of opposite to these angles are denoted by small letters a, b and c, respectively. Sine Rule s i n A a = s i n B b = s i n C c

Cosine Rule a 2  = b 2  + c 2  – 2bc cos A b 2  = c 2  + a 2  – 2ac cos B c 2  = a 2  + b 2  – 2ab cos C

Projection Rule a = b cos C + c cos B b = c cos A + a cos C c = a cos B + b cos A

Trigonometric Functions Class 11 MCQs Questions with Answers

Question 1. The value of cos² x + cos² y – 2cos x × cos y × cos (x + y) is (a) sin (x + y) (b) sin² (x + y) (c) sin³ (x + y) (d) sin 4 (x + y)

Answer: (b) sin² (x + y) Hint: cos² x + cos² y – 2cos x × cos y × cos(x + y) {since cos(x + y) = cos x × cos y – sin x × sin y } = cos² x + cos² y – 2cos x × cos y × (cos x × cos y – sin x × sin y) = cos² x + cos² y – 2cos² x × cos² y + 2cos x × cos y × sin x × sin y = cos² x + cos² y – cos² x × cos² y – cos² x × cos² y + 2cos x × cos y × sin x × sin y = (cos² x – cos² x × cos² y) + (cos² y – cos² x × cos² y) + 2cos x × cos y × sin x × sin y = cos² x(1- cos² y) + cos² y(1 – cos² x) + 2cos x × cos y × sin x × sin y = sin² y × cos² x + sin² x × cos² y + 2cos x × cos y × sin x × sin y (since sin² x + cos² x = 1 ) = sin² x × cos² y + sin² y × cos² x + 2cos x × cos y × sin x × sin y = (sin x × cos y)² + (sin y × cos x)² + 2cos x × cos y × sin x × sin y = (sin x × cos y + sin y × cos x)² = {sin (x + y)}² = sin² (x + y)

Question 2. If a×cos x + b × cos x = c, then the value of (a × sin x – b²cos x)² is (a) a² + b² + c² (b) a² – b² – c² (c) a² – b² + c² (d) a² + b² – c²

Answer: (d) a² + b² – c² Hint: We have (a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b² ⇒ c² + (a × sin x – b × cos x)² = a² + b² ⇒ (a × sin x – b × cos x)² = a² + b² – c²

Question 3. If cos a + 2cos b + cos c = 2 then a, b, c are in (a) 2b = a + c (b) b² = a × c (c) a = b = c (d) None of these

Answer: (a) 2b = a + c Hint: Given, cos A + 2 cos B + cos C = 2 ⇒ cos A + cos C = 2(1 – cos B) ⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2) ⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2) ⇒ cos((A-C)/2) = 2sin (B/2) ⇒ cos((A-C)/2) = 2cos((A+C)/2) ⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2) ⇒ 2sin(A/2)sin(C/2) = sin(B/2) ⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac ⇒ 2(s – b) = b ⇒ a + b + c – 2b = b ⇒ a + c – b = b ⇒ a + c = 2b

Question 4. The value of cos 5π is (a) 0 (b) 1 (c) -1 (d) None of these

Answer: (c) -1 Hint: Given, cos 5π = cos (π + 4π) = cos π = -1

Question 5. In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals (a) none of these (b) c/a (c) 1 (d) a/c

Answer: (c) 1 Hint: Given cosec A (sin B cos C + cos B sin C) = cosec A × sin(B+C) = cosec A × sin(180 – A) = cosec A × sin A = cosec A × 1/cosec A = 1

Question 6. If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is (a) 4 : (√5 – 1) (b) 5 : 4 (c) (√5 – 1) : 4 (d) none of these

Answer: (a) 4 : (√5 – 1) Hint: Given, the angles of a triangle be in the ratio 1 : 4 : 5 ⇒ x + 4x + 5x = 180 ⇒ 10x = 180 ⇒ x = 180/10 ⇒ x = 18 So, the angle are: 18, 72, 90 Since a : b : c = sin A : sin B : sin C ⇒ a : b : c = sin 18 : sin 72 : sin 90 ⇒ a : b : c = (√5 – 1)/4 : {√(10 + 2√5)}/4 : 1 ⇒ a : b : c = (√5 – 1) : {√(10 + 2√5)} : 4 Now, c /a = 4/(√5 – 1) ⇒ c : a = 4 : (√5 – 1)

Question 7. The value of cos 180° is (a) 0 (b) 1 (c) -1 (d) infinite

Answer: (c) -1 Hint: 180 is a standard degree generally we all know their values but if we want to go theoretically then cos(90 + x) = – sin(x) So, cos 180 = cos(90 + 90) = -sin 90 = -1 {sin 90 = 1} So, cos 180 = -1

Question 8. The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is (a) 30° (b) 90° (c) 60° (d) 120°

Answer: (b) 90° Hint: Let the lengths of the sides if ∆ABC be a, b and c Perimeter of the triangle = 2s = a + b + c = 6(sinA + sinB + sinC)/3 ⇒ (sinA + sinB + sinC) = ( a + b + c)/2 ⇒ (sinA + sinB + sinC)/( a + b + c) = 1/2 From sin formula,Using sinA/a = sinB/b = sinC/c = (sinA + sinB + sinC)/(a + b + c) = 1/2 Now, sinB/b = 1/2 Given b = 2 So, sinB/2 = 1/2 ⇒ sinB = 1 ⇒ B = π/2

Question 9: If 3 × tan(x – 15) = tan(x + 15), then the value of x is (a) 30 (b) 45 (c) 60 (d) 90

Answer: (b) 45 Hint: Given, 3×tan(x – 15) = tan(x + 15) ⇒ tan(x + 15)/tan(x – 15) = 3/1 ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = (3 + 1)/(3 – 1) ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 4/2 ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 2 ⇒ sin(x + 15 + x – 15)/sin(x + 15 – x + 15) = 2 ⇒ sin 2x/sin 30 = 2 ⇒ sin 2x/(1/2) = 2 ⇒ 2 × sin 2x = 2 ⇒ sin 2x = 1 ⇒ sin 2x = sin 90 ⇒ 2x = 90 ⇒ x = 45

Question 10. If the sides of a triangle are 13, 7, 8 the greatest angle of the triangle is (a) π/3 (b) π/2 (c) 2π/3 (d) 3π/2

Answer: (c) 2π/3 Hint: Given, the sides of a triangle are 13, 7, 8 Since greatest side has greatest angle, Now Cos A = (b² + c² – a²)/2bc ⇒ Cos A = (7² + 8² – 13²)/(2×7×8) ⇒ Cos A = (49 + 64 – 169)/(2×7×8) ⇒ Cos A = (113 – 169)/(2×7×8) ⇒ Cos A = -56/(2×56) ⇒ Cos A = -1/2 ⇒ Cos A = Cos 2π/3 ⇒ A = 2π/3 So, the greatest angle is = 2π/3

Question 11. The value of tan 20 × tan 40 × tan 80 is (a) tan 30 (b) tan 60 (c) 2 tan 30 (d) 2 tan 60

Answer: (b) tan 60 Hint: Given, tan 20 × tan 40 × tan 80 = tan 40 × tan 80 × tan 20 = [{sin 40 × sin 80}/{cos 40 × cos 80}] × (sin 20/cos 20) = [{2 * sin 40 × sin 80}/{2 × cos 40 × cos 80}] × (sin 20/cos 20) = [{cos 40 – cos 120}/{cos 120 + cos 40}] × (sin 20/cos 20) = [{cos 40 – cos (90 + 30)}/{cos (90 + 30) + cos 40}] × (sin 20/cos 20) = [{cos 40 + sin30}/{-sin30 + cos 40}] × (sin 20/cos 20) = [{(2 × cos 40 + 1)/2}/{(-1 + cos 40)/2}] × (sin 20/cos 20) = [{2 × cos 40 + 1}/{-1 + cos 40}] × (sin 20/cos 20) = [{2 × cos 40 × sin 20 + sin 20}/{-cos 20 + cos 40 × cos 20}] = (sin 60 – sin 20 + sin 20)/(-cos 20 + cos 60 + cos 20) = sin 60/cos 60 = tan 60 So, tan 20 × tan 40 × tan 80 = tan 60

Question 12. If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is (a) 4 : (√5 – 1) (b) 5 : 4 (c) (√5 – 1) : 4 (d) none of these

Question 13. The general solution of √3 cos x – sin x = 1 is (a) x = n × π + (-1)n × (π/6) (b) x = π/3 – n × π + (-1)n × (π/6) (c) x = π/3 + n × π + (-1)n × (π/6) (d) x = π/3 – n × π + (π/6)

Answer: (c) x = π/3 + n × π + (-1)n × (π/6) Hint: √3 cos x-sin x=1 ⇒ (√3/2)cos x – (1/2)sin x = 1/2 ⇒ sin 60 × cos x – cos 60 × sin x = 1/2 ⇒ sin (x – 60) = 1/2 ⇒ sin (x – π/3) = sin 30 ⇒ sin (x – π/3) = sinπ/6 ⇒ x – π/3 = n × π + (-1)n × (π/6) {where n ∈ Z} ⇒ x = π/3 + n × π + (-1)n × (π/6)

Question 14. If tan² θ = 1 – e², then the value of sec θ + tan³ θ × cosec θ is (a) 2 – e² (b) (2 – e²) 1/2 (c) (2 – e²)² (d) (2 – e²) 3/2

Question 15. The value of cos 20 + 2sin² 55 – √2 sin65 is (a) 0 (b) 1 (c) -1 (d) None of these

Answer: (b) 1 Hint: Given, cos 20 + 2sin² 55 – √2 sin65 = cos 20 + 1 – cos 110 – √2 sin65 {since cos 2x = 1 – 2sin² x} = 1 + cos 20 – cos 110 – √2 sin65 = 1 – 2 × sin {(20 + 110)/2 × sin{(20 – 110)/2} – √2 sin65 {Apply cos C – cos D formula} = 1 – 2 × sin 65 × sin (-45) – √2 sin65 = 1 + 2 × sin 65 × sin 45 – √2 sin65 = 1 + (2 × sin 65)/√2 – √2 sin65 = 1 + √2 ( sin 65 – √2 sin 65 = 1 So, cos 20 + 2sin² 55 – √2 sin65 = 1

Question 16. If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is (a) 2π/3 (b) π/3 (c) π/2 (d) π/6

Answer: (a) 2π/3 Hint: Let S be the center of the circumcircle of triangle PQR. So, SP = SQ = SR = PQ = PR, where SP, SQ & SR are radii. Thus SPQ & SPR are equilateral triangles. ⇒ ∠QSP = 60°; Similarly ∠RQP = 60° ⇒ Angle at the center QSP = 120° So, SRPQ is a rhombus, since all the four sides are equal. Hence, its opposite angles are equal; so ∠P = ∠QSP = 120°

Question 17. If cos a + 2cos b + cos c = 2 then a, b, c are in (a) 2b = a + c (b) b² = a × c (c) a = b = c (d) None of these

Answer: (a) 2b = a + c Hint: Given, cos A + 2 cos B + cos C = 2 ⇒ cos A + cos C = 2(1 – cos B) ⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin² (B/2) ⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2) ⇒ cos((A-C)/2) = 2sin (B/2) ⇒ cos((A-C)/2) = 2cos((A+C)/2) ⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2) ⇒ 2sin(A/2)sin(C/2) = sin(B/2) ⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac ⇒ 2(s – b) = b ⇒ a + b + c – 2b = b ⇒ a + c – b = b ⇒ a + c = 2b

Question 18. The value of 4 × sin x × sin(x + π/3) × sin(x + 2π/3) is (a) sin x (b) sin 2x (c) sin 3x (d) sin 4x

Answer: (c) sin 3x Hint: Given, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = 4 × sin x × {sin x × cos π/3 + cos x × sin π/3} × {sin x × cos 2π/3 + cos x × sin 2π/3} = 4 × sin x × {(sin x)/2 + (√3 × cos x)/2} × {-(sin x)/2 + (√3 × cos x)/2} = 4 × sin x × {-(sin 2x)/4 + (3 × cos 2x)/4} = sin x × {-sin 2x + 3 × cos 2x} = sin x × {-sin 2x + 3 × (1 – sin 2x)} = sin x × {-sin 2x + 3 – 3 × sin 2x} = sin x × {3 – 4 × sin 2x} = 3 × sin x – 4 sin 3x = sin 3x So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x

Question 19. If tan A – tan B = x and cot B – cot A = y, then the value of cot (A – B) is (a) x + y (b) 1/x + y (c) x + 1/y (d) 1/x + 1/y

Answer: (d) 1/x + 1/y Hint: Given, tan A – tan B = x ……………. 1 and cot B – cot A = y ……………. 2 From equation, 1/cot A – 1/cot B = x ⇒ (cot B – cot A)/(cot A × cot B) = x ⇒ y/(cot A × cot B) = x {from equation 2} ⇒ y = x × (cot A × cot B) ⇒ cot A × cot B = y/x Now, cot (A – B) = (cot A × cot B + 1)/(cot B – cot A) ⇒ cot (A – B) = (y/x + 1)/y ⇒ cot (A – B) = (y/x) × (1/y) + 1/y ⇒ cot (A – B) = 1/x + 1/y

Question 20. The value of (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) is (a) tan 6x (b) 2 tan 6x (c) 3 tan 6x (d) 4 tan 6x

Answer: (b) 2 tan 6x Hint: Given, (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) ⇒ [{2 × sin(7x+5x)/2 × cos(7x-5x)/2}/{2 × cos(7x+5x)/2 × cos(7x-5x)/2}] + [{2 × sin(9x+3x)/2 × cos(9x-3x)/2}/{2 × cos(9x+3x)/2 × cos(9x-3x)/2}] ⇒ [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] + [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] ⇒ (sin 6x/cos 6x) + (sin 6x/cos 6x) ⇒ tan 6x + tan 6x ⇒ 2 tan 6x

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are available at BYJU’S, which are prepared by our expert teachers. All these solutions are written as per the latest update on the CBSE Syllabus 2023-24 and its guidelines. BYJU’S provides step-by-step solutions by considering the different understanding levels of students and the marking scheme. Chapter 3, Trigonometric Functions, comes under NCERT Class 11 Maths and is an important chapter for students. Though the chapter has numerous mathematical terms and formulae, BYJU’S has made NCERT Solutions for Class 11 Maths easy for the students to understand and remember with the usage of tricks.

Trigonometry has been developed to solve geometric problems involving triangles. Students cannot skip Trigonometric chapters since this is used in many areas, such as finding the heights of tides in the ocean, designing electronic circuits, etc. In the earlier classes, students have already studied trigonometric identities and applications of trigonometric ratios. In Class 11 , trigonometric ratios are generalised to trigonometric function and their properties. However, the NCERT Solutions of BYJU’S help the students to attain more knowledge and score full marks in this chapter of the examination.

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

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In this section, a few important terms are defined, such as principal solution and general solution of trigonometric functions, which are explained using examples too. Exercise 3.1 Solutions 7 Questions Exercise 3.2 Solutions 10 Questions Exercise 3.3 Solutions 25 Questions Exercise 3.4 Solutions 9 Questions Miscellaneous Exercise On Chapter 3 Solutions 10 Questions

Access NCERT Solutions for Class 11 Maths Chapter 3

Exercise 3.1 page: 54

1. Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°

2. Find the degree measures corresponding to the following radian measures (Use π = 22/7)

Here π radian = 180°

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

It is given that

No. of revolutions made by the wheel in

1 minute = 360

1 second = 360/60 = 6

We know that

The wheel turns an angle of 2π radian in one complete revolution.

In 6 complete revolutions, it will turn an angle of 6 × 2π radian = 12 π radian

Therefore, in one second, the wheel turns an angle of 12π radian.

4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7).

5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

The dimensions of the circle are

Diameter = 40 cm

Radius = 40/2 = 20 cm

Consider AB be as the chord of the circle i.e. length = 20 cm

Radius of circle = OA = OB = 20 cm

Similarly AB = 20 cm

Hence, ΔOAB is an equilateral triangle.

θ = 60° = π/3 radian

In a circle of radius  r  unit, if an arc of length  l  unit subtends an angle  θ  radian at the centre

We get θ = 1/r

Therefore, the length of the minor arc of the chord is 20π/3 cm.

6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

7. Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length

(i) 10 cm (ii) 15 cm (iii) 21 cm

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = 1/r

We know that r = 75 cm

(i) l = 10 cm

θ = 10/75 radian

By further simplification

θ = 2/15 radian

(ii) l = 15 cm

θ = 15/75 radian

θ = 1/5 radian

(iii) l = 21 cm

θ = 21/75 radian

θ = 7/25 radian

Exercise 3.2 page: 63

Find the values of other five trigonometric functions in Exercises 1 to 5.

1. cos x = -1/2, x lies in third quadrant.

2. sin x = 3/5, x lies in second quadrant.

sin x = 3/5

We can write it as

sin 2 x + cos 2 x = 1

cos 2 x = 1 – sin 2 x

3. cot x = 3/4, x lies in third quadrant.

cot x = 3/4

1 + tan 2 x = sec 2 x

1 + (4/3) 2 = sec 2 x

Substituting the values

1 + 16/9 = sec 2 x

cos 2 x = 25/9

sec x = ± 5/3

Here x lies in the third quadrant so the value of sec x will be negative

sec x = – 5/3

4. sec x = 13/5, x lies in fourth quadrant.

sec x = 13/5

sin 2 x = 1 – cos 2 x

sin 2 x = 1 – (5/13) 2

sin 2 x = 1 – 25/169 = 144/169

sin 2 x = ± 12/13

Here x lies in the fourth quadrant so the value of sin x will be negative

sin x = – 12/13

5. tan x = -5/12, x lies in second quadrant.

tan x = – 5/12

1 + (-5/12) 2 = sec 2 x

1 + 25/144 = sec 2 x

sec 2 x = 169/144

sec x = ± 13/12

Here x lies in the second quadrant so the value of sec x will be negative

sec x = – 13/12

Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765°

We know that values of sin x repeat after an interval of 2π or 360°

By further calculation

7. cosec (–1410°)

We know that values of cosec x repeat after an interval of 2π or 360°

= cosec 30 o = 2

We know that values of tan x repeat after an interval of π or 180°

Exercise 3.3 page: 73

Prove that:

= 1/2 + 4/4

5. Find the value of:

(i) sin 75 o

(ii) tan 15 o

(ii) tan 15°

It can be written as

= tan (45° – 30°)

Using formula

Prove the following:

= sin x cos x (tan x + cot x)

10. sin ( n  + 1) x  sin ( n  + 2) x  + cos ( n  + 1) x  cos ( n  + 2) x  = cos  x

LHS = sin ( n  + 1) x  sin ( n  + 2) x  + cos ( n  + 1) x  cos ( n  + 2) x

Using the formula

12. sin 2  6 x  – sin 2  4 x  = sin 2 x  sin 10 x

13. cos 2  2 x  – cos 2  6 x  = sin 4 x  sin 8 x

= (2 sin 4 x  cos 4 x ) (2 sin 2 x  cos 2 x )

= sin 8x sin 4x

14. sin 2x + 2sin 4x + sin 6x = 4cos 2  x sin 4x

= 2 sin 4x cos (– 2x) + 2 sin 4x

= 2 sin 4x cos 2x + 2 sin 4x

Taking common terms

= 2 sin 4x (cos 2x + 1)

= 2 sin 4x (2 cos 2  x – 1 + 1)

= 2 sin 4x (2 cos 2  x)

= 4cos 2  x sin 4x

15. cot 4 x  (sin 5 x  + sin 3 x ) = cot  x  (sin 5 x  – sin 3 x )

LHS = cot 4x (sin 5x + sin 3x)

= 2 cos 4x cos x

Hence, LHS = RHS.

22. cot  x  cot 2 x  – cot 2 x  cot 3 x  – cot 3 x  cot  x  = 1

LHS = tan 4x = tan 2(2x)

By using the formula

24. cos 4 x  = 1 – 8sin 2  x  cos 2  x

LHS = cos 4x

= cos 2(2 x )

Using the formula cos 2 A  = 1 – 2 sin 2   A

= 1 – 2 sin 2  2 x

Again by using the formula sin2 A  = 2sin  A  cos A

= 1 – 2(2 sin  x  cos  x ) 2

= 1 – 8 sin 2 x  cos 2 x

25. cos 6 x  = 32 cos 6   x  – 48 cos 4   x  + 18 cos 2   x  – 1

L.H.S. = cos 6 x

= cos 3(2 x )

Using the formula cos 3 A  = 4 cos 3   A  – 3 cos   A

= 4 cos 3  2 x  – 3 cos   2 x

Again by using formula cos 2 x  = 2 cos 2   x  – 1

= 4 [(2 cos 2   x  – 1) 3  – 3 (2 cos 2   x  – 1)

= 4 [(2 cos 2   x ) 3  – (1) 3  – 3 (2 cos 2   x ) 2  + 3 (2 cos 2   x )] – 6cos 2   x  + 3

= 4 [8cos 6 x  – 1 – 12 cos 4 x  + 6 cos 2 x ] – 6 cos 2 x  + 3

By multiplication

= 32 cos 6 x  – 4 – 48 cos 4 x  + 24 cos 2   x  – 6 cos 2 x  + 3

On further calculation

= 32 cos 6 x  – 48 cos 4 x  + 18 cos 2 x  – 1

Exercise 3.4 PAGE: 78

Find the principal and general solutions of the following equations:

1. tan x = √3

2. sec x = 2

3. cot x = – √3

4. cosec x = – 2

Find the general solution for each of the following equations:

5. cos 4x = cos 2x

6. cos 3x + cos x – cos 2x = 0

7. sin 2x + cos x = 0

sin 2x + cos x = 0

2 sin x cos x + cos x = 0

cos x (2 sin x + 1) = 0

cos x = 0 or 2 sin x + 1 = 0

Let cos x = 0

8. sec 2 2x = 1 – tan 2x

sec 2 2x = 1 – tan 2x

1 + tan 2 2x = 1 – tan 2x

tan 2 2x + tan 2x = 0

tan 2x (tan 2x + 1) = 0

tan 2x = 0 or tan 2x + 1 = 0

If tan 2x = 0

tan 2x = tan 0

2x = nπ + 0, where n ∈ Z

x = nπ/2, where n ∈ Z

tan 2x + 1 = 0

tan 2x = – 1

2x = nπ + 3π/4, where n ∈ Z

x = nπ/2 + 3π/8, where n ∈ Z

Hence, the general solution is nπ/2 or nπ/2 + 3π/8, n ∈ Z.

9. sin x + sin 3x + sin 5x = 0

sin x + sin 3x + sin 5x = 0

(sin x + sin 5x) + sin 3x = 0

2 sin 3x cos (-2x) + sin 3x = 0

2 sin 3x cos 2x + sin 3x = 0

By taking out the common terms

sin 3x (2 cos 2x + 1) = 0

sin 3x = 0 or 2 cos 2x + 1 = 0

If sin 3x = 0

3x = nπ, where n ∈ Z

x = nπ/3, where n ∈ Z

If 2 cos 2x + 1 = 0

cos 2x = – 1/2

= – cos π/3

= cos (π – π/3)

cos 2x = cos 2π/3

Miscellaneous Exercise page: 81

2. (sin 3 x  + sin  x ) sin  x  + (cos 3 x  – cos  x ) cos  x  = 0

LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

= sin 3x sin x + sin 2 x + cos 3x cos x – cos 2 x

Taking out the common terms

= cos 3x cos x + sin 3x sin x – (cos 2 x – sin 2 x)

cos (A – B) = cos A cos B + sin A sin B

= cos (3x – x) – cos 2x

= cos 2x – cos 2x

LHS = (cos x + cos y) 2 + (sin x – sin y) 2

By expanding using formula we get

= cos 2 x + cos 2 y + 2 cos x cos y + sin 2 x + sin 2 y – 2 sin x sin y

Grouping the terms

= (cos 2 x + sin 2 x) + (cos 2 y + sin 2 y) + 2 (cos x cos y – sin x sin y)

Using the formula cos (A + B) = (cos A cos B – sin A sin B)

= 1 + 1 + 2 cos (x + y)

= 2 + 2 cos (x + y)

Taking 2 as common

From the formula cos 2A = 2 cos 2 A – 1

LHS = (cos x – cos y) 2 + (sin x – sin y) 2

By expanding using formula

= cos 2 x + cos 2 y – 2 cos x cos y + sin 2 x + sin 2 y – 2 sin x sin y

= (cos 2 x + sin 2 x) + (cos 2 y + sin 2 y) – 2 (cos x cos y + sin x sin y)

Using the formula cos (A – B) = cos A cos B + sin A sin B

From formula cos 2A = 1 – 2 sin 2 A

5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

8. Find sin x/2, cos x/2 and tan x/2 in each of the following:

cos x = -3/5

From the formula

9. cos x = -1/3, x in quadrant III

10. sin x = 1/4, x in quadrant II

This chapter has 6 exercises and a miscellaneous exercise to help students understand the concepts related to Trigonometric Functions clearly. BYJU’S provides all the concepts and solutions for Chapter 3 of Class 11 Maths with clear explanations and formulas. The PDF of Maths  NCERT Solutions for Class 11 Chapter 3 includes the topics and sub-topics listed below. 3.1 Introduction The basic trigonometric ratios and identities are given here, along with the applications of trigonometric ratios in solving the word problems related to heights and distances. 3.2 Angles 3.2.1 Degree measure 3.2.2 Radian measure 3.2.3 Relation between radian and real numbers 3.2.4 Relation between degree and radian In this section, different terms related to trigonometry are discussed, such as terminal side, initial sides, measuring an angle in degrees and radian, etc. 3.3 Trigonometric Functions 3.3.1 Sign of trigonometric functions 3.3.2 Domain and range of trigonometric functions After studying this section, students are able to understand the generalised trigonometric functions with signs. Also, they can gain knowledge on domain and range of trigonometric functions with examples. 3.4 Trigonometric Functions of Sum and Difference of Two Angles This section contains formulas related to the sum and difference of two angles in trigonometric functions.

Key Features of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Studying the Trigonometric Functions of Class 11 enables the students to understand the following:

• Introduction to Trigonometric Functions
• Positive and negative angles
• Measuring angles in radians and in degrees and conversion of one into other
• Definition of trigonometric functions with the help of unit circle
• Truth of the sin 2x + cos 2x = 1, for all x
• Signs of trigonometric functions
• Domain and range of trigonometric functions
• Graphs of Trigonometric Functions
• Expressing sin (x±y) and cos (x±y) in terms of sin x, sin y, cos x & cosy and their simple application, identities related to sin 2x, cos 2x, tan 2x, sin 3x, cos 3x and tan 3x
• The general solution of trigonometric equations of the type sin y = sin a, cos y = cos a and tan y = tan a

Disclaimer – 

Dropped Topics – 

3.5 Trigonometric Equations (up to Exercise 3.4) Last five points in the Summary 3.6 Proofs and Simple Applications of Sine and Cosine Formulae

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Syllabus for the Session 2023-24

CBSE Syllabus

Case Study Questions

Case Study on Sets      CS-2   CS-3   CS-4   CS-5

Case Study on Relations & Functions   CS-2   CS-3

Case Study on Trigonometric Functions

Case Study on Complex Numbers

Case Study on Linear Inequalities

Case Study on Permutations and Combinations

Case Study on Sequences & Series

Case Study on Straight Lines

Case Study on Conic Sections

Case Study on Statistics

Case Study on Probability

Pdf of  Case Studies

MCQs for Practice

Chapter 1 - Sets     

Chapter 2 - Relations & Functions

Chapter 3 - Trigonometric Functions

Chapter 4 - Complex Numbers & Quadratic Equations

Chapter 5 - Linear Inequalities

Chapter 6 - Permutations & Combinations

Chapter 7 - Binomial Theorem

Chapter 8 - Sequences & Series

Chapter 9 - Straight Line

Chapter 1 0 - Conic Sections

Chapter 1 1 - Introduction to Three-dimensional Geometry

Chapter 12 - Limits & D erivatives

Chapter 13 - Statistics

Chapter 14 - Probability

Answers of MCQs

Assertion & Reasoning Questions

Relations & Functions

Trigonometric Functions

Complex Numbers

Linear Inequalities

Permutations & Combinations

Binomial Theorem

Sequences & Series

Straight Lines

Conic Sections

Introduction to Three-Dimensional Geometry

Limits & derivatives

Probability

Topic Wise Assignments of Previous Year Questions

Relations and Functions

Straight Line

Limits & Derivatives

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            SA2(QP)         SA2(MS)         COMP(QP)         COMP(MS)

Session 201 6 -20 17

            SA1(QP)      SA2(QP)         SA2(MS)         COMP(QP)        COMP(MS)

Session 201 7 -201 8

            SA2(QP)          SA2(MS)         COMP(QP)         COMP(MS)

Session 201 8 -201 9

            SA2(QP)          SA2(MS)         COMP(QP)         COMP(MS)

Session 201 9 -20 20

            SA2(QP)          SA2(MS)        

Session 2020-21

    Video Explanation:  Part A     Part B Section III     Part B Section IV   Part B Section V

Session 2021-22

Question Paper 

Video Explanation:  Section A     Section B     Section C

Session 2022-23

  Sample Question Paper for Mid-Term Exam

Video Explanation: Section A     Section B    Section C     Section D     Section E

Mid-Term Exam:  Question Paper

Video Explanation: Section A     Section B     Section C     Section D    Section E

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Video Explanation: Section A     Section B     Section C     Section D     Section E

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Video Explanation: Section A     Section B     Section C     Section D     Section E

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Short Capsules (Notes to Revise Concepts)

Chapter 1 - Sets

Chapter 4 - Mathematical Induction

Chapter 5 - Complex Numbers & Quadratic Equations

Chapter 6 - Linear Inequalities

Chapter 7 - Permutations & Combinations

Chapter 8 - Binomial Theorem

Chapter 9 - Sequences & Series

Chapter 10 - Straight Line

Chapter 11 - Conic Sections

Chapter 12 - Introduction to Three-dimensional Geometry

Chapter 13 - Limits & derivatives

Chapter 14 - Mathematical Reasoning

Chapter 15 - Statistics

Chapter 16 - Probability

Some Basic Concepts to Revise:

Number System

Mensuration Results

Quadratic Equations

Solution of Polynomial Inequality - Wavy Curve Method

Probability__Revision of the Topics studied in Earlier Classes  

Result Sheets

Trigonometric Identities

NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

September 22, 2019 by phani

Get Free NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1, Ex 3.2, Ex 3.3, Ex 3.4, and Miscellaneous Exercise PDF in Hindi and English Medium. Trigonometric Functions Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. Trigonometric Functions All Exercises Class 11 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers.

Class 11 Maths Trigonometric Functions NCERT Solutions in English Medium and Hindi Medium

• Trigonometric Functions Class 11 Ex 3.1
• Trigonometric Functions Class 11 Ex 3.2
• Trigonometric Functions Class 11 Ex 3.3
• Trigonometric Functions Class 11 Ex 3.4
• Trigonometric Functions Class 11 Miscellaneous Exercise

त्रिकोणमितीय फलन प्रश्नावली 3.1 का हल हिंदी में

• त्रिकोणमितीय फलन प्रश्नावली 3.2 का हल हिंदी में
• त्रिकोणमितीय फलन प्रश्नावली 3.3 का हल हिंदी में
• त्रिकोणमितीय फलन प्रश्नावली 3.4 का हल हिंदी में
• त्रिकोणमितीय फलन विविध प्रश्नावली का हल हिंदी में
• Trigonometry Formulas
• Trigonometry Functions Class 11 Notes
• NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions
• Trig Cheat Sheet
• JEE Main Trigonometry Previous Year Questions

Free download NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1, Ex 3.2, Ex 3.3, Ex 3.4, and Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.1

Ex 3.1 Class 11 Maths Question 1: Find the radian measures corresponding to the following degree measures: (i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520° Ans:

(i) \(\frac{11}{16}\) We know that: π radian = 180° ∴ \(\frac{11}{16}\) radain = \(\frac{180}{\pi} \times \frac{11}{16}\) × degree

= \(\frac{45 \times 11}{\pi \times 4}\) degree

= \(\frac{45 \times 11 \times 7}{22 \times 4}\) degree

= \(\frac{315}{8}\) degree

= 39 \(\frac{3}{8}\) degree

= 39° + \(\frac{3 \times 60}{8}\) minutes [1° = 60′]

= 39° + 22′ + \(\frac{1}{2}\) minutes

= 39°22’30” [1′ = 60°].

More Resources for CBSE Class 11

NCERT Solutions

• NCERT Solutions Class 11 Maths
• NCERT Solutions Class 11 Physics
• NCERT Solutions Class 11 Chemistry
• NCERT Solutions Class 11 Biology
• NCERT Solutions Class 11 Hindi
• NCERT Solutions Class 11 English
• NCERT Solutions Class 11 Business Studies
• NCERT Solutions Class 11 Accountancy
• NCERT Solutions Class 11 Psychology
• NCERT Solutions Class 11 Entrepreneurship
• NCERT Solutions Class 11 Indian Economic Development
• NCERT Solutions Class 11 Computer Science

Ex 3.1 Class 11 Maths Question 3: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? Ans: Number of revolutions made by the wheel in 1 minute = 360 ∴ Number of revolutions made by the wheel in 1 second = \(\frac{360}{6}\) = 6 In one complete revolution, the wheel turns an angle of 2π radian. Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e., 12π radian Thus, in one second, the wheel turns an angle of 12π radian. Ex 3.1 Class 11 Maths Question 4: Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm y an arc of length 22 cm (Use π = \(\frac{22}{7}\)). Ans:

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = \(\frac{l}{r}\) Therefore, for r = 100 cm, l = 22 cm, we have

Thus, the required angle is 12°36′.

Ex 3.1 Class 11 Maths Question 5: In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord. Ans:

Given, diameter = 40 cm ∴ radius (r) = \(\frac{40}{2}\) = 20 cm and length of chord, AB = 20 cm Thus, ∆OAB is an equilateral triangle. We know that, θ = \(\frac{\text { Arc } A B}{\text { radius }}\) ⇒ Arc AB = θ × r = \(\frac{\pi}{3}\) × 20 . = \(\frac{20}{3}\) π cm.

Ex 3.1 Class 11 Maths Question 6: If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii. Ans:

Let the radii of the two circles be r 1  and r 2 . Let an arc of length l subtend an angle of 60° at the centre of the circle of radius r 1 , while let an arc of length l subtend an angle of 75° at the centre of the circle of radius r 2 . Now, 6o° = \(\frac{\pi}{3}\) radian and 75° = \(\frac{5 \pi}{12}\) radian We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = \(\frac{l}{r}\) or l = rθ ∴ l = \(\frac{r_{1} \pi}{3}\) and

l = \(\frac{r_{2} 5 \pi}{12}\)

⇒ \(\frac{r_{1} \pi}{3}=\frac{r_{2} 5 \pi}{12}\)

⇒ r = \(\frac{r_{2} 5}{4}\)

\(\frac{r_{1}}{r_{2}}=\frac{5}{4}\) Thus, the ratio of the radii is 5 : 4.

Ex 3.1 Class 11 Maths Question 7: Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm. Ans:

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = \(\frac{l}{r}\). It is given that r = 75 cm

(i) Here, l = 10 cm θ = \(\frac{10}{75}\) radian = \(\frac{2}{15}\) radian

(ii) Here, l = 15 cm θ = \(\frac{15}{75}\) radian θ = \(\frac{1}{5}\) radian

(iii) Here, l = 21 cm θ = \(\frac{21}{75}\) radian = \(\frac{7}{75}\) radian.

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.2

sin x = \(\frac{3}{5}\)

cosec x = \(\frac{1}{\sin x}=\frac{1}{\left(\frac{3}{5}\right)}=\frac{5}{3}\) sin 2  x + cos 2  x = 1 ⇒ cos 2  x = 1 – sin 2  x ⇒ cos 2  x = 1 – (\(\frac{3}{5}\)) 2

⇒ cos 2  x = 1 – \(\frac{9}{25}\)

⇒ cos 2  x = \(\frac{16}{25}\)

⇒ cos x = ± \(\frac{4}{5}\) Since x lies in the 2nd quadrant, the value of cos x will be negative

⇒ \(\frac{4}{3}=\frac{\sin x}{\frac{-3}{5}}\) ⇒ sin x = \(\left(\frac{4}{3}\right) \times\left(\frac{-3}{5}\right)=-\frac{4}{5}\) ⇒ cosec x = \(\frac{1}{\sin x}=-\frac{5}{4}\).

tan x = – \(\frac{5}{12}\)

cot x = \(\frac{1}{\tan x}=\frac{1}{\left(-\frac{5}{12}\right)}=-\frac{12}{5}\)

Ex 3.2 Class 11 Maths Question 6: Find the value of the trigonometric function sin 765°. Ans: It is known that the values of sin x repeat after an interval of 2π or 360°. ∴ sin 765° = sin (2 × 360° + 45°) = sin 45° = 1 Ex 3.2 Class 11 Maths Question 7: Find the value of the trigonometric function cosec (- 1410°) Ans: It is known that the values of cosec x repeat after an interval of 2π or 360°. ∴ cosec (- 1410°) = cosec (- 1410° + 4 x 360°) = cosec (- 1410° + 1440°) = cosec 30° = 2. Ex 3.2 Class 11 Maths Question 8: Find the value of the trigonometric function tan \(\frac{19 \pi}{3}\). Ans:

It is known that the values of tan x repeat after an interval of π or 180°. ∴ \(\tan \frac{19 \pi}{3}=\tan 6 \frac{1}{3} \pi\)

= \(\tan \left(6 \pi+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}\)

= tan 60° = √3.

Ex 3.2 Class 11 Maths Question 9: Find the value of the trigonometric function sin \(\left(-\frac{11 \pi}{3}\right)\). Ans:

It is known that the values of cot x repeat after an interval of π or 180°.

∴ \(\sin \left(\frac{11 \pi}{3}\right)=\sin \left(-\frac{11 \pi}{3}+2 \times 2 \pi\right)\)

= \(\sin \left(\frac{\pi}{3}\right)=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\)

Ex 3.2 Class 11 Maths Question 10: Find the value of the trigonometric function cot \(\left(-\frac{15 \pi}{4}\right)\). Ans:

It is known that the values of cot x repeat after an interval of ir or 1800. ∴ \(\cot \left(-\frac{15 \pi}{4}\right)=\cot \left(-\frac{15 \pi}{4}+4 \pi\right)=\cot \frac{\pi}{4}\) = 1.

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.3

Ex 3.3 Class 11 Maths Question 1:

Prove that: sin 2  \(\frac{\pi}{6}\) + cos 2  \(\frac{\pi}{3}\) – tan 2  \(\frac{\pi}{4}\) = – \(\frac{1}{2}\) Ans:

L.H.S.= sin 2  \(\frac{\pi}{6}\) + cos 2  \(\frac{\pi}{3}\) – tan 2  \(\frac{\pi}{4}\)

= \(\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\) – (1) 2

= \(\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}\)

= R.H.S. Hence proved.

L.H.S = \(2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}\)

= \(2\left\{\sin \left(\pi-\frac{\pi}{4}\right)\right\}^{2}+2\left(\frac{1}{\sqrt{2}}\right)^{2}+2(2)^{2}\)

= \(2\left\{\sin \frac{\pi}{4}\right\}^{2}+2 \times \frac{1}{2}+8\)

= 2 \(\left(\frac{1}{\sqrt{2}}\right)^{2}\) + 1 + 8

= 1 + 1 + 8 = 10 = R.H.S. Hence proved.

Ex 3.3 Class 11 Maths Question 5: Find the value of: (i) sin 75°, (ii) tan 15° Ans:

(i) sin 75° sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° [∵ sin (x + y) = sin x cos y + cos x sin y] = \(\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\)

= \(\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

(ii) tan 15° = tan (45° – 30°)

L.H.S = \(\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}\)

= \(\frac{[-\cos x][\cos x]}{(\sin x)(-\sin x)}=\frac{-\cos ^{2} x}{-\sin ^{2} x}\)

= R.H.S Hence proved.

It is known that cos A – cos B = \(-2 \sin \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right)\)

∴ L.H.S.= \(=\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)\)

= \(– 2 \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)+\left(\frac{3 \pi}{4}-x\right)}{2}\right\} \cdot \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)-\left(\frac{3 \pi}{4}-x\right)}{2}\right\}\)

= – 2 sin (\(\frac{3 \pi}{4}\)) sin x

= – 2 sin (- \(\frac{\pi}{4}\)) sin x

= – √2 sin x = R.H.S. Hence proved.

Ex 3.3 Class 11 Maths Question 12: Prove that: sin 2  6x – sin 2  4x = sin 2x sin 10 x Ans:

It is known that sin A + sin B = 2 \(\sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

sin A – sin B = 2 \(\cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\) L.H.S.= sin 2  6x – sin 2  4x = (sin 6x + sin 4x) (sin 6x – sin 4x) = (2 sin 5x cos x) (2 cos 5x sin x) = (2 sin 5x cos 5x) (2 sin x cos x) = sin 10x sin 2x = R.H.S. Hence proved.

Ex 3.3 Class 11 Maths Question 13: Prove that: cos 2  2x cos 2  6x = sin 4x sin 8x Ans:

It is known that cos A + cos B = 2 \(\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

cos A – cos = 2 \(\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\)

∴ L.H.S = cos 2  2x – cos 2  6x = (cos 2x + cos 6x) (cos 2x – 6x) = \(\left[2 \cos \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]\left[-2 \sin \left(\frac{2 x+6 x}{2}\right) \sin \frac{(2 x-6 x)}{2}\right]\) ∴ L.H.S.= cos 2  2x – cos 2  6x = (cos 2x + cos 6x) (cos 2x – 6x) = [2 cos 4x cos (-2x)] [- 2 sin 4x sin (- 2x)] = [2 cos 4x cos 2x] [- 2 sin 4x (- sin 2x)] = (2 sin 4x cos 4x) (2 sin 2x cos 2x) = sin 8x sin 4x = R.H.S. Hence proved.

It is known that sin A + sin = 2 \(\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

cos A + cos = 2 \(\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

∴ L.H.S = \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\)

= \(\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}\)

= \(\frac{2 \sin 4 x \cos x}{2 \cos 4 x \cos x}=\frac{\sin 4 x}{\cos 4 x}\)

= tan 4x = R.H.S. Hence proved.

It is known that sin A – sin B = 2 \(\cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\)

cos 2  A – sin 2  A = cos 2A

∴ L.H.S. = \(=\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}\)

= \(\frac{2 \cos \left(\frac{x+3 x}{2}\right) \sin \left(\frac{x-3 x}{2}\right)}{-\cos 2 x}\)

= \(\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}\)

= – 2 × (- sin x) = 2 sin x

Hence proved.

Class 11 Maths NCERT Solutions Chapter 3 Exercise 3.4

cos 4x = cos 2x cos 4x – cos 2x = 0 – 2 sin \(\left(\frac{4 x+2 x}{2}\right)\) sin \(\left(\frac{4 x-2 x}{2}\right)\) = 0

[∵ cos A – cos B = 2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\(\)]

sin 3x sin x = 0 sin 3x = 0or sin x = 0 3x = nπ or x = nπ, where n ∈ Z x = \(\frac{n \pi}{3}\) or x = nπ, where n ∈ Z.

Ex 3.4 Class 11 Maths Question 6: Find the general solution of the equation cos 3x + cosx – cos 2x = 0 Ans:

cos 3x + cos x – cos 2x = 0 2 cos \(\left(\frac{3 x+x}{2}\right)\) cos \(\left(\frac{3 x-x}{2}\right)\) – cos 2x = 0

[∵ cos A + cos B = 2 \(\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)]

2 cos 2x cos x – cos 2x = 0 cos 2x (2 cos x – 1) = 0 cos 2x = 0 or 2 cos x – 1 = 0 cos 2x = 0 or cos x = \(\frac{1{2}\) ∴ 2x = (2n + 1) \(\frac{\pi}{2}\) or cos x = cos \(\frac{\pi}{3}\), where n ∈ Z x = (2n + 1) \(\frac{\pi}{4}\) or x = 2nπ ± \(\frac{\pi}{3}\) where n ∈ Z.

Ex 3.4 Class 11 Maths Question 7: Find the general solution of the equation sin 2x + cos x = 0 Ans:

sin 2x + cos x = 0 ⇒ 2sin x cos x + cos x = 0 ⇒ cos x (2 sin x + 1) = 0 ⇒ cos x = 0 or 2 sin x + 1 = 0 Now, cos x = 0 ⇒ x = (2n + 1) \(\frac{\pi}{2}\) , where n ∈ Z. or 2 sin x + 1 = 0 ⇒ sin x = – \(\frac{1}{2}\)

= – sin \(\frac{\pi}{6}\)

= sin (π + \(\frac{\pi}{6}\))

= sin \(\frac{7 \pi}{6}\)

x = nπ + (- 1) n  \(\frac{7 \pi}{6}\) where n ∈ Z Therefore, the general solution is (2n + 1) \(\frac{\pi}{2}\) or nπ + (- 1) n  \(\frac{7 \pi}{6}\) where n ∈ Z.

Ex 3.4 Class 11 Maths Question 8: Find the general solution of the equation sec 2  2x = 1 – tan 2x. Ans:

sec 2  2x = 1 – tan 2x 1 + tan 2  2x = 1 – tan 2x tan 2  x + tan 2x = 0 => tan 2x (tan 2x + 1) = 0 => tan 2x = 0 or tan 2x + 1 = 0 Now, tan 2x = 0 => tan 2x = tan 0 2x = nπ + 0, where n ∈ Z x = \(\frac{n \pi}{2}\), where n ∈ Z or tan 2x + 1 = 0 = tan 2x = – 1 = – tan \(\frac{\pi}{4}\)

= tan (π – \(\frac{\pi}{4}\))

= tan \(\frac{3 \pi}{4}\)

2x = nπ + \(\frac{3 \pi}{4}\) where n ∈ Z

x = \(\frac{n \pi}{2}+\frac{3 \pi}{8}\), where n ∈ Z

Therefore, the general solution is \(\frac{n \pi}{2}\) or \(\frac{n \pi}{2}+\frac{3 \pi}{8}\) where n ∈ Z.

Ex 3.4 Class 11 Maths Question 9: Find the general solution of the equation sin x + sin 3x + sin 5x = 0 Ans:

sin x + sin 3x + sin 5x = 0 ⇒ (sin x + sin 5x) + sin 3x = 0

\(\left[2 \sin \left(\frac{x+5 x}{2}\right) \cos \left(\frac{x-5 x}{2}\right)\right]\) + sin 3x = 0

[∵ sin A + sin B = 2 sin \(\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)]

2 sin 3x cos (2x) + sin 3x = 0 2 sin 3x cos 2x + sin 3x = 0 sin 3x (2 cos 2x +1) = 0 sin 3x = 0 or 2 cos 2x + 1 = 0 Now sin 3x = 0 ⇒ 3x = nπ, where n ∈ Z i.e., x = \(\frac{n \pi}{3}\) where n ∈ Z or 2 cos 2x + 1 = 0 cos 2x = \(-\frac{1}{2}\)

= – cos \(\frac{\pi}{3}\)

= cos (π – \(\frac{\pi}{3}\))

cos 2x = cos \(\frac{2 \pi}{3}\)

⇒ 2x = 2nπ ± \(\frac{2\pi}{3}\), where n ∈ Z

⇒ x = nπ ± \(\frac{\pi}{3}\), where n ∈ Z

Therefore, the general solution is \(\frac{n \pi}{3}\) or nπ ± \(\frac{\pi}{3}\), where n ∈ Z.

NCERT Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercise

L.H.S. = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = sin 3x sin x + sin 2  x + cos 3x cos x – cos 2  x = cos 3x cos x + sin 3x sin x – (cos 2  x – sin 2  x) = cos (3x – x) – cos 2x [∵ cos(A – B) = cos A cos B + sin A sin B] = cos 2x – cos 2x = 0 =R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 3:

Prove that: (cos x + cos y) 2  + (sin x – sin y) 2  = 4 cos 2  \(\left(\frac{x+y}{2}\right)\)

L.H.S.= (cos x + cos y) 2  + (sin x – sin y) 2 = cos 2  x + cos 2  y + 2 cos x cos y + sin 2  x + sin 2  y – 2 sin x sin y = (cos 2  x + sin 2  x) + (cos 2  y + sin 2  y) + 2 (cos x cos y – sin x sin y) = 1 + 1 + 2 cos (x + y) [∵ cos (A + B) = (cos A cos B – sin A sin B)] = 2 + 2 cos (x + y) = 2 [1 + cos (x + y)] = 2[1 + \(2 \cos ^{2}\left(\frac{x+y}{2}\right)\) – 1] [∵ cos 2A = 2 cos 2  A – 1] = 4 c0s 2  \(\left(\frac{x+y}{2}\right)\) = R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 4: Prove that: (cos x – cos y) 2  + (sin x – sin y) 2  = 4 sin 2  \(\frac{x-y}{2}\) Ans:

L.H.S.= (cos x – cos y) 2  + (sin x – sin y) 2 = cos 2  x + cos 2  y – 2 cos x cos y + sin 2  x + sin 2  y – 2 sin x sin y = (cos 2  x + sin 2  x) + (cos 2  y + sin 2  y) – 2 [cos x cos y + sin x sin y] = 1 + 1 – 2 [cos (x – y)] = 2 [1 – {1 – 2 sin 2  \(\left(\frac{x-y}{2}\right)\)}] [∵ cos 2A = 1 – 2 sin 2  A] = 4 sin 2  \(\left(\frac{x-y}{2}\right)\) = R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 5: Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x Ans:

It is known that sin A + sin B = 2 \(\sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)\) ∴ L.H.S. = (sin x + sin 3x) + (sin 5x + sin 7x) = (sin x + sin 5x) + (sin 3x + sin 7x) = \(2 \sin \left(\frac{x+5 x}{2}\right)\) . \(\cos \left(\frac{x-5 x}{2}\right)+2 \sin \left(\frac{3 x+7 x}{2}\right) \cos \left(\frac{3 x-7 x}{2}\right)\) = 2 sin 3x cos (- 2x) + 2 sin 5x cos (- 2x) = 2 sin 3x cos 2x + 2 sin 5x cos 2x = 2 cos 2x [sin 3x + sin 5x] = 2 cos 2x [latex]2 \sin \left(\frac{3 x+5 x}{2}\right) \cdot \cos \left(\frac{3 x-5 x}{2}\right)[/latex] = 2 cos 2x [2 sin 4x . cos (- x)] = 4 cos 2x sin 4x cos x = R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 6: Prove that: \(\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}\) = tan 6x Ans: It is known that

प्रश्न 3. एक-पहिया एक मिनट में 360° परिक्रमण करता है तो एक सेकंड में कितने रेडियन माप का कोण बनाएगा? हल: परिक्रमण में पहिया द्वारा बना कोण = 27 रेडियन 360 परिक्रमण में पहिया द्वारा बना कोण = 360 x 2π रेडियन 1 मिनट अर्थात् 60 सेकण्ड में 360 x 2π रेडियन का कोण बनता है। 1 सेकण्ट में चहिया द्वारा बना कोण = \(\frac { 360\times 2\pi }{ 60 }\) = 12π रेडियन।

Exercise 3.1

Q.1: Calculate the radian measurement of the given degree measurement:

(ii). 240 ∘

(iii). − 47 ∘ 30 ‘

(iv). 520 ∘

Q.2: Calculate the degree measurement of the given degree measurement: [Use π = \(\\ \frac { 22 }{ 7 } \)]

(i) \(\\ \frac { 11 }{ 16 } \)

(iii) \(\frac { 5\pi }{ 3 } \)

(iv) \(\frac { 7\pi }{ 6 } \)

Q.3: In a minute, wheel makes 360 revolutions. Through how many radians does it turn in 1 second?

Q.4: Calculate the degree measurement of the angle subtended at the centre of a circle of radius 100 m by an arc of length 22 m.

Q.5: In a circle of diameter 40 m, the length of the chord 20 m. Find the length of minor arc of chord.

Q.6: In two circles, arcs which has same length subtended at an angle of 60 ∘ and 75 ∘ at the center. Calculate the ratio of their radii.

Q.7: Calculate the angle in radian through which a pendulum swings if the length is 75 cm and the tip describes an arc of length

(iii) 21 cm

Exercise 3.2

Q.1: Calculate the values of five trigonometric func. if cos y = \(– \frac { 1 }{ 2 } \) and y lies in 3 rd quadrant.

(iii) cosec y

Q.2: Calculate the other five trigonometric function if we are given the values for sin y = \(\\ \frac { 3 }{ 5 } \), where y lies in second quadrant.

Q.3: Find the values of other five trigonometric functions if c o t y =\(\\ \frac { 3 }{ 4 } \) , where y lies in the third quadrant.

Q.4: Find the values of other five trigonometric if s e c y =\(\\ \frac { 13 }{ 5 } \) , where y lies in the fourth quadrant.

Q.5: Find the values of other five trigonometric function if tan y = \(– \frac { 5 }{ 12 } \) and y lies in second quadrant.

Q.6: Calculate the value of trigonometric function sin 765°.

Q.7: Calculate the value of trigonometric function cosec [-1410°]

Q.8: Calculate the value of the trigonometric function tan \(\frac { 19\pi }{ 3 } \) .

Q.9: Calculate the value of the trigonometric function sin \(-\frac { 11\pi }{ 3 } \) .

Q.10: Calculate the value of the trigonometric function cot \(-\frac { 15\pi }{ 4 } \)

Exercise 3.3

Q.1: Prove:

sin²\(\frac { \pi }{ 6 } \) + cos² \(\frac { \pi }{ 3 } \) – tan² \(\frac { \pi }{ 4 } \) = \(– \frac { 1 }{ 2 } \)

Q.2: Prove:

2 sin² \(\frac { \pi }{ 6 } \) + c o s e c ² \(\frac { 7\pi }{ 6 } \) 6 cos ² \(\frac { \pi }{ 3 } \) =\(\\ \frac { 3 }{ 2 } \)

Q.3: Prove:

cot ² \(\frac { \pi }{ 6 } \) + c o s e c \(\frac { 5\pi }{ 6 } \) + 3 tan ² latex s=2]\frac { \pi }{ 6 } [/latex] = 6

Q.4: Prove:

2 sin ² \(\frac { 3\pi }{ 4 } \) + 2 cos ² \(\frac { \pi }{ 4 } \) + 2 sec ² \(\frac { \pi }{ 3 } \) = 10

Q.5: Calculate the value of:

(i). sin 75 ∘

(ii). tan 15 ∘

cos ( \(\frac { \pi }{ 4 } \) – x ) cos ( \(\frac { \pi }{ 4 } \) – y ) – sin ( \(\frac { \pi }{ 4 } \) – x ) sin ( \(\frac { \pi }{ 4 } \) – y ) = sin ( x + y )

Q.7: Prove:

\(\frac { tan(\frac { \pi }{ 4 } +x) }{ tan(\frac { \pi }{ 4 } -x) } ={ \left( \frac { 1+tanx }{ 1-tanx } \right) }^{ 2 }\)

Q.8: Prove:

\(\frac { cos(\pi +x)cos(-x) }{ sin(\pi -x)cos\left( \frac { \pi }{ 2 } +x \right) } ={ cot }^{ 2 }x\)

Q.9: Prove:

\(cos(\frac { 3\pi }{ 2 } +x)cos(2\pi +x)[cot(\frac { 3\pi }{ 2 } -x)+cot(2\pi +x)]=1\)

Q.10: Prove:

sin ( n + 1 ) x sin ( n + 2 ) x + cos ( n + 1 ) x cos ( n + 2 ) x = cos x

Q.11 Prove:

\(cos(\frac { 3\pi }{ 4 } +x)-cos(\frac { 3\pi }{ 4 } -x)\)= − √2 sin x

Q.12: Prove:

sin² 6 x – sin ² 4 x = sin2 x sin 10 x

Q.13: Prove:

cos ² 2 x – cos ² 6 x = sin 4 x sin 8 x

Q.14:Prove:

sin 2 x + 2 sin 4 x + sin 6 x = 4 cos ² x sin 4 x

Q.15: Prove:

cot 4 x ( sin 5 x + sin 3 x ) = cot x ( sin 5 x – sin 3 x )

Q.16: Prove:

\(\frac { cos9x-cos5x }{ sin17x-sin3x } =-\frac { sin2x }{ cos10x } \)

Q.17: Prove:

\(\frac { sin5x+sin3x }{ cos5x+cos3x } =tan4x\)

Q.18: Prove:

\(\frac { sinx-siny }{ cosx+cosy } =tan\frac { x-y }{ 2 } \)

Q.19: Prove:

\(\frac { sinx+sin3x }{ cosx+cos3x } =tan2x\)

Q.20: Prove:

\(\frac { sinx-sin3x }{ { sin }^{ 2 }x-{ cos }^{ 2 }x } =2sinx\)

Q.21: Prove:

\(\frac { cos4x+cos3x+cos2x }{ sin4x+sin3x+sin2x } =cot3x\)

Q.22: Prove:

cot x cot 2 x – cot 2 x cot 3 x – cot 3 x cot x = 1

Q.23: Prove:

\(tan4x=\frac { 4tanx(1-{ tan }^{ 2 }x) }{ 1-6{ tan }^{ 2 }x+{ tan }^{ 4 }x } \)

Q.24: Prove:

cos 4 x = 1 – 8 sin² x cos² x

Q.25: Prove:

cos 6 x = 32 cos 6 x – 48 cos 4 x + 18 cos 2 x − 1

Exercise 3.4

Q.1: Find general solutions and the principle solutions of the given equation: tan x = √3

Q.2: Find general solutions and the principle solutions of the given equation: sec x = 2

Q.3: Find general solutions and the principle solutions of the given equation: cot = − √3

Q.4: Find general solutions and the principle solutions of the given equation: cosec x = -2

Q.5: Find the general solution of the given equation: cos 4x = cos 2x

Q.6: Find the general solution of the given equation: cos 3x + cos x – cos 2x = 0

Q.7: Find the general solution of the given equation: sin 2x + cos x = 0

Q.8: Find the general solution of the given equation: sec² 2 x = 1 – tan 2 x

Q.9: Find the general solution of the given equation: sin x + sin 3x + sin 5x = 0

Miscellaneous Exercise

Q.1: Prove that:

\(2cos\frac { \pi }{ 13 } cos\frac { 9\pi }{ 13 } +cos\frac { 3\pi }{ 13 } +cos\frac { 5\pi }{ 13 } =0\)

Q.2: Prove that:

( sin 3 x + sin x ) sin x + ( cos 3 x – cos x ) cos x = 0

Q-3: Prove that:

( cos x + cos y )² + ( sin x – sin y ) ² = 4 cos ²\(\\ \frac { x+y }{ 2 } \)

Q-4: Prove that:

( cos x – cos y ) ² + ( sin x – sin y ) ² = 4 sin ² \(\\ \frac { x-y }{ 2 } \)

Q-5: Prove that:

sin x + sin 3 x + sin 5 x + sin 7 x = 4 cos x cos 2 x cos 4 x

Q-6: Prove that:

\(\frac { (sin7x+sin5x)+(sin9x+sin3x) }{ (cos7x+cos5x)+(cos9x+cos3x) } =tan6x\)

Q-7: Show that: sin 3 y + sin 2 y – sin y = 4 sin y cos\(\\ \frac { y }{ 2 } \) cos\(\\ \frac { 3y }{ 2 } \)

Q-8: The value of tan y =\(– \frac { 4 }{ 2 } \) where y in in 2 nd quadrant then find out the values of sin\(\\ \frac { y }{ 2 } \) , cos\(\\ \frac { y }{ 2 } \) a n d tan\(\\ \frac { y }{ 2 } \) .

Q-9: The value of cos y = \(– \frac { 1 }{ 3 } \) where y in in 3 rd quadrant then find out the values of sin \(\\ \frac { y }{ 2 } \) , cos \(\\ \frac { y }{ 2 } \) a n d tan \(\\ \frac { y }{ 2 } \) .

Q-10: The value of sin y = \(\\ \frac { 1 }{ 4 } \) where y in in 2 nd quadrant then find out the values of sin \(\\ \frac { y }{ 2 } \) , cos \(\\ \frac { y }{ 2 } \) a n d tan \(\\ \frac { y }{ 2 } \) .

NCERT Solutions for Class 11 Maths All Chapters

• Chapter 1 Sets
• Chapter 2 Relations and Functions
• Chapter 3 Trigonometric Functions
• Chapter 4 Principle of Mathematical Induction
• Chapter 5 Complex Numbers and Quadratic Equations
• Chapter 6 Linear Inequalities
• Chapter 7 Permutation and Combinations
• Chapter 8 Binomial Theorem
• Chapter 9 Sequences and Series
• Chapter 10 Straight Lines
• Chapter 11 Conic Sections
• Chapter 12 Introduction to Three Dimensional Geometry
• Chapter 13 Limits and Derivatives
• Chapter 14 Mathematical Reasoning
• Chapter 15 Statistics
• Chapter 16 Probability

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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions PDF Download

NCERT Solutions are essential for all students who want to score well in their exams. Our subject matter experts have developed these solutions to help the students score good marks in their examinations. These solutions are well-explained and include each and every topic. Another essential benefit of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions is that it is created as per the latest pattern of CBSE (Central Board of Secondary Education). These solutions also ensure that a student gets a clear understanding of each and every concept. 

In the Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Solutions, each and every type of question is included. From easy to hard, the students will get an idea of all the types of questions which can appear in the exam. Also, apart from this, the students will also get to know about the most common questions which have a high chance of appearing in the exam. 

It can be assumed that a student will score good marks in their exams if they will study the Maths Class 11 Chapter 3 Trigonometric Functions PDF NCERT Solution on a regular basis. Our subject matter experts at Selfstudys have invested a good amount of time in creating them so that the students do not find any difficulty studying from these solutions. As the experts have been in the education line for quite some time, they are aware of the learning potential of every student, which is why they have created these solutions in a simple way so that no student faces any difficulty studying from them. Students can also use these solutions at the time of revision. 

The NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions can be easily downloaded from the website of Selfstudys i.e. Selfstudys.com. 

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions PDF 

The NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions PDF can be easily downloaded from the website of Selfstudys. As they are in PDF format, it becomes easier for the students to access them. They are also mobile-friendly. 

How to Download the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions? 

Downloading the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions is if you know the right steps. The steps to download them are as follows: 

• Visit the official website of Selfstudys i.e. Selfstudys.com.

• Once the website is completely loaded, you need to click on the three lines which you will find on the upper left side. Following that, you have to click on the ‘NCERT Books and Solutions’ column. 

• After clicking on the option of ‘NCERT Books and Solutions’, a new list will open where you need to select the option of ‘NCERT Solutions’. 

• Following that, you will be redirected to a page where you need to select the class and the subject for which you want to download the NCERT Solutions. 
• In this case, you need to select class 10 and the subject Maths to download the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions. 

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There are many features of the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions which can be very beneficial for all the students. The most important features include: 

• Well-Elaborated: The NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions is well-elaborated to make it easier for the students to understand all the concepts with in-depth knowledge. 
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• Accuracy and reliability: Our subject matter experts have invested a good amount of time in developing these NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions. They are accurate and reliable for the preparation of exams. These notes can also be helpful for the students who do self-study. 

What are the Benefits of the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions? 

The NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions has several benefits to offer to the students. Some of the most important of them are 

• Clear Understanding of the Topic: The Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Solutions provides in-depth knowledge of each and every topic which is present in the syllabus. Each chapter is explained in detail in a step-by-step process. To make learning easier and more interesting for all the students, diagrams and examples are also given. 
• Authentic Solutions: The Maths Class 11 Chapter 3 Trigonometric Functions PDF NCERT Solution are completely authentic and can be trusted 100%. The highly qualified subject matter experts at Selfstudys have created these solutions in an authentic way which makes it simple for all the students to learn all the concepts. 
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• Conceptual Clarity: The NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions focuses on building a strong foundation for all the students. Students can get a deep understanding of the fundamental principles of maths formulas, which serve as a basis for further mathematical topics. 
• Boosts Confidence: If a student regularly goes through the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions, there is no doubt that the students will gain confidence in their maths power. This confidence can motivate them to explore more difficult topics and make a strong base for all of them. 

What are the tips to study from the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions? 

There are various tips that can help the students to study effectively from the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions: 

• Go through the chapter thoroughly: It is very important for all the students to first go through the chapter as it will help them to understand the basics. If their basics will get strong, they will automatically be able to grasp the other concepts in a better way. 
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• Take note of your mistakes and work on them: It is advisable for all the students to take note of their mistakes while solving the Maths Class 11 Chapter 3 Trigonometric Functions PDF NCERT Solution and work on them. Find your weaknesses and identify the specific concepts or steps that caused confusion. This will help you improve your problem-solving skills. 

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To maximize your learning potential while solving the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions, here are the tips which can help you: 

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Studying for the exam requires focus and concentration. Below we will discuss various methods of how students can study for an Exam with the help of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions: 

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• Study the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions on a regular basis: It is advisable for all the students to study Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Solutions on a regular basis as it will help them to do effective preparation for their exams and also increases the chances of the students to score good marks in their exams. 
• Always read the summary of your textbook: It is advisable for all students to always read the summary which is included in the Maths Class 11 Chapter 3 Trigonometric Functions PDF NCERT Solution. The students should read them on a regular basis as it will help them to get a thorough knowledge of all the topics and will also help them to prepare well for the exams. This will increase their chances of scoring good marks in the exams. 
• Identify your strong and weak areas: It is very important for all the students to identify their strong and weak areas as it will help them to understand where they need to put more effort in order to do well in their exams. 
• Do timely revision: All the students should do timely revision from the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions if they want to do extremely well in their exams as it will ensure that you have created a strong base of all the topics in their minds. 

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Unit 3: Trigonometric functions

Degrees and radians.

• Intro to radians (Opens a modal)
• Radians & degrees (Opens a modal)
• Degrees to radians (Opens a modal)
• Radians to degrees (Opens a modal)
• Radian angles & quadrants (Opens a modal)
• Radians & degrees Get 3 of 4 questions to level up!

Unit circle

• Unit circle (Opens a modal)
• The trig functions & right triangle trig ratios (Opens a modal)
• Trig values of π/4 (Opens a modal)
• Trig unit circle review (Opens a modal)
• Plotting angles on unit circle Get 3 of 4 questions to level up!
• Sign of trigonometric functions Get 3 of 4 questions to level up!
• Trig values of special angles Get 3 of 4 questions to level up!

Trigonometric functions

• Graph of y=sin(x) (Opens a modal)
• Graph of y=tan(x) (Opens a modal)
• Intersection points of y=sin(x) and y=cos(x) (Opens a modal)
• Find value of other trigonometric functions from given trigonometric function Get 3 of 4 questions to level up!

Trigonometric identities: Symmetry

• Sine & cosine identities: symmetry (Opens a modal)
• Tangent identities: symmetry (Opens a modal)
• Sine & cosine identities: periodicity (Opens a modal)
• Tangent identities: periodicity (Opens a modal)

Trigonometric identities: Sum and difference

• Trig angle addition identities (Opens a modal)
• Using the cosine angle addition identity (Opens a modal)
• Using the cosine double-angle identity (Opens a modal)
• Proof of the sine angle addition identity (Opens a modal)
• Proof of the cosine angle addition identity (Opens a modal)
• Trig challenge problem: cosine of angle-sum (Opens a modal)
• Trigonometric functions of sum and difference of angles Get 3 of 4 questions to level up!
• Sum and difference of trigonometric functions Get 3 of 4 questions to level up!
• Evaluate trigonometric expressions (intermediate) Get 3 of 4 questions to level up!

Trigonometric equations

• Proof of the Pythagorean trig identity (Opens a modal)
• Using the Pythagorean trig identity (Opens a modal)
• Solving sinusoidal equations of the form sin(x)=d (Opens a modal)
• Solving cos(θ)=1 and cos(θ)=-1 (Opens a modal)
• Use the Pythagorean identity Get 3 of 4 questions to level up!
• Principal solutions of trigonometric equation Get 3 of 4 questions to level up!
• General solution of trigonometric equation Get 3 of 4 questions to level up!

Solutions to select NCERT problems

• Select problems from exercise 3.3 (Opens a modal)
• Select problems from miscellaneous exercise (Opens a modal)

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Assertion Reason Questions for Class 11 Maths Chapter 3 Trigonometric Functions

• Last modified on: 1 year ago
• Reading Time: 12 Minutes

[PDF] Download Assertion Reason Questions for Class 11 Maths Chapter 3 Trigonometric Functions

Here we are providing assertion reason questions for class 11 maths. In this article, we are covering Class 11 Maths Chapter 3 Trigonometric Functions Assertion Reason Questions. Detailed Solutions are also provided at the end of questions so that students can check their answers after completing all the questions.

Related Posts

What is assertion reason questions.

Assertion Reason Questions are a type of question that can be asked in Class 11 Maths exams. These questions usually consist of two parts:

• Assertion: A statement that presents a claim or proposition that needs to be verified or evaluated.
• Reason: A statement that provides a logical explanation or justification for the assertion.

The student is required to evaluate both the assertion and the reason and determine whether they are true or false. Here are some examples of Assertion Reason Questions in Class 11 Maths:

Example 1: Assertion: The sum of the angles of a triangle is 180 degrees. Reason: The angles of a triangle are in a ratio of 1:2:3.

Solution: The assertion is true as it is a well-known fact in geometry that the sum of the angles of a triangle is 180 degrees. However, the reason is false as the angles of a triangle are not necessarily in the ratio of 1:2:3. Therefore, the correct option is (a) Assertion is true, but the reason is false.

Example 2: Assertion: The equation x^2 + 1 = 0 has no real solutions. Reason: The square of any real number is always non-negative.

Solution: The assertion is true as x 2 + 1 = 0 has no real solutions. The reason is also true as the square of any real number is always non-negative. Therefore, the correct option is (c) Assertion is true, and the reason is also true, and the reason is a correct explanation for the assertion.

Importance of Practicing Assertion Reason Questions for Class 11 Maths

Practicing Assertion Reason Questions in Class 11 Maths has several benefits:

• Develops critical thinking skills: Assertion Reason Questions require students to analyze and evaluate statements logically. By practicing such questions, students can develop their critical thinking skills and learn to reason effectively.
• Improves problem-solving ability: Assertion Reason Questions often involve complex concepts and require students to apply their knowledge to solve problems. Practicing such questions can help students improve their problem-solving ability and build a strong foundation in Maths.
• Helps in exam preparation: Assertion Reason Questions are frequently asked in Class 11 Maths exams. By practicing such questions regularly, students can get familiar with the exam pattern and improve their chances of scoring well in the exam.
• Enhances conceptual understanding: Assertion Reason Questions involve analyzing the relationship between concepts, which can help students understand the underlying principles better. By practicing such questions, students can improve their conceptual understanding of Maths.
• Builds confidence: Regular practice of Assertion Reason Questions can help students build confidence in their ability to solve complex problems and reason effectively. This can help them perform better in exams and in their future academic and professional pursuits.

Why CBSE Students Fear Assertion Reason Questions?

CBSE students may fear assertion-reasoning questions because they require a deeper level of understanding and analytical skills compared to regular multiple-choice questions. Assertion-reasoning questions consist of two statements: an assertion and a reason. The student must determine if both statements are true and if the reason explains the assertion.

These questions require the student to not only know the facts but also understand the relationships between them. Students may fear these types of questions if they are not confident in their ability to analyze and evaluate the information provided. Additionally, these questions often carry a good weightage, which can add to the pressure of performing well on the exam.

However, with practice and a solid understanding of the concepts, students can overcome their fear of assertion-reasoning questions and perform well on their exams. It is important for students to read the questions carefully, understand the meaning of each statement, and analyze the relationship between the two statements before selecting their answer.

What are the best ways to prepare for assertion reason questions?

Preparing for assertion-reasoning questions can be challenging, but with the right approach and strategies, you can improve your skills and perform well on exams. Here are some ways to prepare for assertion-reasoning questions:

• Understand the concepts: To answer assertion-reasoning questions correctly, you need to have a clear understanding of the concepts involved. Make sure you study the topic thoroughly and have a strong grasp of the fundamental principles.
• Practice regularly: Practice is key to improving your skills. Look for sample assertion-reasoning questions in textbooks or online resources and practice answering them. Try to identify the relationship between the two statements and determine if both are true and if the reason explains the assertion.
• Develop analytical skills: Assertion-reasoning questions require you to analyze and evaluate the information provided. Develop your analytical skills by reading and analyzing different types of texts, such as news articles or scientific research papers.
• Take notes: Taking notes while studying can help you remember important information and make connections between different concepts. Write down key points, definitions, and examples to refer to later.
• Seek help: If you are struggling with assertion-reasoning questions, don’t hesitate to seek help from your teacher or a tutor. They can provide you with additional resources and guidance on how to approach these types of questions.

By following these strategies and putting in the effort to practice, you can improve your skills and perform well on assertion-reasoning questions.

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Case Study Questions for Class 11 Maths Chapter 3 Trigonometric Functions

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[PDF] Download Case Study Questions for Class 11 Maths Chapter 3 Trigonometric Functions

Here we are providing case study questions for class 11 maths. In this article, we are sharing Class 11 Maths Chapter 3 Trigonometric Functions. All case study questions of class 11 maths are solved so that students can check their solutions after attempting questions.

What is meant by Case Study Question?

In the context of CBSE (Central Board of Secondary Education), a case study question is a type of question that requires students to analyze a given scenario or situation and apply their knowledge and skills to solve a problem or answer a question related to the case study.

Case study questions typically involve a real-world situation that requires students to identify the problem or issue, analyze the relevant information, and apply their understanding of the relevant concepts to propose a solution or answer a question. These questions may involve multiple steps and require students to think critically, apply their problem-solving skills, and communicate their reasoning effectively.

Importance of Solving Case Study Questions for Class 11 Maths

Case study questions are an important aspect of mathematics education at the Class 11 level. These questions require students to apply their knowledge and skills to real-world scenarios, helping them develop critical thinking, problem-solving, and analytical skills. Here are some reasons why case study questions are important in Class 11 maths education:

• Real-world application: Case study questions allow students to see how the concepts they are learning in mathematics can be applied in real-life situations. This helps students understand the relevance and importance of mathematics in their daily lives.
• Higher-order thinking: Case study questions require students to think critically, analyze data, and make connections between different concepts. This helps develop higher-order thinking skills, which are essential for success in both academics and real-life situations.
• Collaborative learning: Case study questions often require students to work in groups, which promotes collaborative learning and helps students develop communication and teamwork skills.
• Problem-solving skills: Case study questions require students to apply their knowledge and skills to solve complex problems. This helps develop problem-solving skills, which are essential in many careers and in everyday life.
• Exam preparation: Case study questions are included in exams and tests, so practicing them can help students prepare for these assessments.

Overall, case study questions are an important component of Class 11 mathematics education, as they help students develop critical thinking, problem-solving, and analytical skills, which are essential for success in both academics and real-life situations.

Feature of Case Study Questions on This Website

Here are some features of a Class 11 Maths Case Study Questions Booklet:

Many Case Study Questions: This website contains many case study questions, each with a unique scenario and problem statement.

Different types of problems: The booklet includes different types of problems, such as optimization problems, application problems, and interpretation problems, to test students’ understanding of various mathematical concepts and their ability to apply them to real-world situations.

Multiple-choice questions: Questions contains multiple-choice questions to assess students’ knowledge, understanding, and critical thinking skills.

Focus on problem-solving skills: The questions are designed to test students’ problem-solving skills, requiring them to identify the problem, select appropriate mathematical tools, and analyze and interpret the results.

Emphasis on practical applications: The case studies in the booklet focus on practical applications of mathematical concepts, allowing students to develop an understanding of how mathematics is used in real-life situations.

Comprehensive answer key: The booklet includes a comprehensive answer key that provides detailed explanations and step-by-step solutions for all the questions, helping students to understand the concepts and methods used to solve each problem.

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• Class 11 Mathematics Case...

Class 11 Mathematics Case Study Questions

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

If you’re seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations. Furthermore, it is routinely updated to bring you up to speed with the newest CBSE syllabus. So, why delay? Begin your path to success with myCBSEguide now!

The rationale behind teaching Mathematics

The general rationale to teach Mathematics at the senior secondary level is to assist students:

• In knowledge acquisition and cognitive understanding of basic ideas, words, principles, symbols, and mastery of underlying processes and abilities, notably through motivation and visualization.
• To experience the flow of arguments while demonstrating a point or addressing an issue.
• To use the information and skills gained to address issues using several methods wherever possible.
• To cultivate a good mentality in order to think, evaluate, and explain coherently.
• To spark interest in the subject by taking part in relevant tournaments.
• To familiarise pupils with many areas of mathematics utilized in daily life.
• To pique students’ interest in studying mathematics as a discipline.

Case studies in Class 11 Mathematics

A case study in mathematics is a comprehensive examination of a specific mathematical topic or scenario. Case studies are frequently used to investigate the link between theory and practise, as well as the connections between different fields of mathematics. A case study will frequently focus on a specific topic or circumstance and will investigate it using a range of methodologies. These approaches may incorporate algebraic, geometric, and/or statistical analysis.

Sample Class 11 Mathematics case study questions

When it comes to preparing for Class 11 Mathematics, one of the best things Class 11 Mathematics students can do is to look at some Class 11 Mathematics sample case study questions. Class 11 Mathematics sample case study questions will give you a good idea of the types of Class 11 Mathematics sample case study questions that will be asked in the exam and help you to prepare more effectively.

Looking at sample questions is also a good way to identify any areas of weakness in your knowledge. If you find that you struggle with a particular topic, you can then focus your revision on that area.

myCBSEguide offers ample Class 11 Mathematics case study questions, so there is no excuse. With a little bit of preparation, Class 11 Mathematics students can boost their chances of getting the grade they deserve.

Some samples of Class 11 Mathematics case study questions are as follows:

Class 11 Mathematics case study question 1

• 9 km and 13 km
• 9.8 km and 13.8 km
• 9.5 km and 13.5 km
• 10 km and 14 km
• x  ≤   −1913
• x <  −1613
• −1613  < x <  −1913
• There are no solution.
• y  ≤   12 x+2
• y >  12 x+2
• y  ≥   12 x+2
• y <  12 x+2

Answer Key:

• (b) 9.8 km and 13.8 km
• (a) −1913   ≤  x 
• (b)  y >  12 x+2
• (d) (-5, 5)

Class 11 Mathematics case study question 2

• 2 C 1 × 13 C 10
• 2 C 1 × 10 C 13
• 1 C 2 × 13 C 10
• 2 C 10 × 13 C 10
• 6 C 2​ × 3 C 4   × 11 C 5 ​
• 6 C 2​ × 3 C 4   × 11 C 5
• 6 C 2​ × 3 C 5 × 11 C 4 ​
• 6 C 2 ​  ×   3 C 1 ​  × 11 C 5 ​
• (b) (13) 4  ways
• (c) 2860 ways.

Class 11 Mathematics case study question 3

Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the rate of Rs 1000 /yard². The builder planned ground floor of 5 m height, first floor of 4.75 m and so on each floor is 0.25 m less than its previous floor.

Class 11 Mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

• 5x + 6y = 60
• 6x + 5y = 60
• (a) (10, 0)
• (b) 6x + 5y = 60
• (b) 60/√ 61 km
• (d) 2√61 km

A peek at the Class 11 Mathematics curriculum

The Mathematics Syllabus has evolved over time in response to the subject’s expansion and developing societal requirements. The Senior Secondary stage serves as a springboard for students to pursue higher academic education in Mathematics or professional subjects such as Engineering, Physical and Biological Science, Commerce, or Computer Applications. The current updated curriculum has been prepared in compliance with the National Curriculum Framework 2005 and the instructions provided by the Focus Group on Teaching Mathematics 2005 in order to satisfy the rising demands of all student groups. Greater focus has been placed on the application of various principles by motivating the themes from real-life events and other subject areas.

Class 11 Mathematics (Code No. 041)

Design of Class 11 Mathematics exam paper

CBSE Class 11 mathematics question paper is designed to assess students’ understanding of the subject’s essential concepts. Class 11 mathematics question paper will assess their problem-solving and analytical abilities. Before beginning their test preparations, students in Class 11 maths should properly review the question paper format. This will assist Class 11 mathematics students in better understanding the paper and achieving optimum scores. Refer to the Class 11 Mathematics question paper design provided.

 Class 11 Mathematics Question Paper Design

• No chapter-wise weightage. Care to be taken to cover all the chapters.
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.  

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections.

  Prescribed Books:

• Mathematics Textbook for Class XI, NCERT Publications
• Mathematics Exemplar Problem for Class XI, Published by NCERT
• Mathematics Lab Manual class XI, published by NCERT

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Chapter 3 Class 11 Trigonometric Functions

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NCERT Solutions of Chapter 3 Class 11 Trigonometry is available free at teachoo. You can check the detailed explanation of all questions of exercises, examples and miscellaneous by clicking on the Exercise link below.

We had learned Basics of Trigonometry in Class 10. In this chapter, we will learn

• What is a positive or a negative angle
• Measuring angles in Degree , Minutes and Seconds
• Radian measure of an angle
• Converting Degree to Radians , and vice-versa
• Sign of sin, cos, tan in all 4 quadrants
• Finding values of trigonometric functions when one value is given (Example: Finding value of sin, cot, cosec, tan, sec, when cos x = -3/5 is given)
• Finding Value of trigonometric functions, given angle
• Solving questions by formula like  (x + y) formula, 2x 3x formula, Cos x + cos y formula , 2 sin x sin y formula 
• Finding principal and general solutions of a trigonometric equation
• Sin and Cosine Formula with supplementary Questions

Important questions are marked, and Formula sheet is also provided. Click on an exercise or topic to begin.

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Important Questions for CBSE Class 11 Maths Chapter 3 – Trigonometric Functions

Important questions for cbse class 11 maths chapter 3 - trigonometric functions, cbse class 11 maths chapter-3 important questions - free pdf download.

Free PDF download of Important Questions with solutions for CBSE Class 11 Maths Chapter 3 - Trigonometric Functions prepared by expert Maths teachers from latest edition of CBSE(NCERT) books. CoolGyan.Org to score more marks in your Examination.

1 Marks Questions

1. Find the radian measure corresponding to 5° 37" 30""

Ans. 39°22"30""

3. Find the length of an arc of a circle of radius 5 cm subtending a central angle measuring 15°

5. Find the value of sin(–1125°)

6. Find the value of tan 15°

9. Express sin 12 θ + sin 4 θ as the product of sines and cosines.

Ans. 2 sin8θ cos4θ

10. Express 2 cos4x sin2x as an algebraic sum of sines or cosines.

Ans. sin 6x – sin2x

11. Write the range of cos θ

Ans. [–1,1]

12. What is domain of sec θ ?

13.Find the principal solutions of cotx = 3 

14. Write the general solution of cos θ = 0

17.Convert into radian measures. – 47 0 30ʹ

18.Evaluate tan 75 0 .

Ans. tan 75 = tan (45 + 30)

Ans. L. H. S = Sin (40 + θ). Cos (10 + θ) – Cos (40 + θ). Sin (10 +θ)

22.Convert into radian measures. -37 0 30’

Ans. L.H.S = Cos (n+1) x Cos (n+2) x + Sin (n+1) x Sin (n+2) x

Ans. L. H. S = tan 36 0

32.Prove Cos 4x = 1 – 8 Sin 2 x. Cos 2 x

Ans. L. H. S = Cos 4x

Ans. L. H. S = tan 56 0

= tan (45 0 + 11 0 )

35.Prove that Cos 105 0 + Cos 15 0 = Sin 75 0 – Sin 15 0

Ans. L. H. S = Cos 105 0 + Cos 15 0

36.Find the value of Cos (- 1710 0 ).

Ans. Cos (-1710 0 ) = Cos (1800-90)[Cos (-θ) = Cos θ

37.A wheel makes 360 revolutions in 1 minute. Through how many radians does it turn in 1 second.

Ans. N. of revolutions made in 60 sec. = 360

38.Prove Sin 2 6x – Sin 2 4x = Sin2 x. Sin 10 x.

Ans. L. H. S = Sin 2 6x – Sin 2 4x

= Sin (6x + 4x). Sin (6x – 4x)

= Sin 10x . Sin 2x

Ans. L. H. S = tan (69 + 66)

= tan (135)

= tan (90 + 45)

4 Marks Questions

Prove the following Identities

1.The minute hand of a watch is 1.5 cm long. How far does it tip move in 40 minute?

Ans. r = 1.5 cm

Angle made in 60 mint = 360 0

2. Show that tan 3x. tan 2x. tan x = tan 3x – tan 2x – tan x

Ans. Let 3x = 2x + x

tan 3x = tan (2x + x)

5.If in two circles, arcs of the same length subtend angles 60 0 and 75 0 at the centre find the ratio of their radii.

6.Prove that Cos 6x= 32 Cos 6 x – 48 Cos 4 x + 18 Cos 2 x-1

Ans. L.H.S. = Cos 6x

7.Solve Sin2x-Sin4x+Sin6x=o

8.In a circle of diameter 40cm, the length of a chord is 20cm. Find the length of minor are of the chord.

Ans. L. H. S = tan 4x

Ans. L. H. S = (Cos x + Cos y) 2 + (Sin x – Sin y) 2

13.Prove that Sin x + Sin 3x + Sin 5x + Sin 7x = 4 Cos x. Cos 2x. Sin 4x

Ans. L. H. S. = Sin x + Sin 3x + Sin 5x + Sin 7x

= Sin x + Sin 7x + Sin 3x + Sin 5x

14.Find the angle between the minute hand and hour hand of a clock when the time is 7. 20.

Angle made by hour hand in 1 hr = 30 0

Angle made = 90 + 10 = 100 0

16.Prove that Cot 4x (Sin 5x + Sin 3x) = Cot x (Sin 5x – Sin 3x)

Ans. L. H. S = Cot 4x (Sin 5x + Sin 3x)

R. H. S = Cot x (Sin 5x – Sin 3x)

6 Marks Questions

1. Find the general solution of sin2x + sin4x + sin6x = 0

4. Prove that Cos α + Cos β + Cos γ + Cos ( α + β + γ )

Ans. L. H. S.

Ans. L.H.S.

7. Find the value of tan ( α + β ) Given that

Ans. L. H. S=

Ans. L. H. S = Cos 20 0 . Cos 40 0 . Cos 60 0 . Cos 80 0 .

= Cos 60. Cos 20 0 . Cos 40 0 . Cos 80.

Cos x is – tive

NCERT Solutions Class 11 Maths Chapter 3 Trigonometric Functions

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are formulated to explain the fundamental application of trigonometric formulas and the graphs of their functions. Trigonometry is an important part of Class 11 maths that involves studying various relations between sides and angles of triangles . Trigonometric Functions are applied to define these relations and have numerous applications in various other fields, including science and engineering. They are often used for basic geometric calculations and to explain numeric solutions. The knowledge of Trigonometric Functions is vital for subjects like astronomy and geography. NCERT Solutions Class 11 Maths Chapter 3 will offer the right foundational skills in students to study trigonometric relations and functions along with their practical applications.

Chapter 3 of Class 11 Maths will enable students to generalize the concept of trigonometric ratios to trigonometric functions. Learning about the properties of Trigonometric Functions and operations based on them is crucial for math studies. With the regular practice of Class 11 Maths NCERT Solutions Chapter 3, students will quickly master the solutions to equations using Trigonometric Functions. The wide range of problems and examples provided in these solutions are beneficial in promoting an in-depth understanding of concepts. To learn and practice with NCERT Solutions Chapter 3 Trigonometric Functions, download the exercises provided in the links below.

• NCERT Solutions Class 11 Maths Chapter 3 Ex 3.1
• NCERT Solutions Class 11 Maths Chapter 3 Ex 3.2
• NCERT Solutions Class 11 Maths Chapter 3 Ex 3.3
• NCERT Solutions Class 11 Maths Chapter 3 Ex 3.4
• NCERT Solutions Class 11 Maths Chapter 3 Miscellaneous Ex

NCERT Solutions for Class 11 Maths Chapter 3 PDF

Trigonometry sees the use of many formulas, theorems, and steps to solve the sums hence, ensuring that kids allot an ample amount of time for practice is very important. NCERT Solutions for Class 11 Maths Chapter 3 are proficiently modeled to support higher-level math learning. The comprehensive format of these solutions is highly reliable to enhance problem-solving skills in engaging ways. To learn and practice trigonometric functions with these solutions, click on the links of the pdf file given below.

☛ Download Class 11 Maths NCERT Solutions Chapter 3 Trigonometric Functions

NCERT Class 11 Maths Chapter 3   Download PDF

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

NCERT solutions Class 11 Maths Chapter 3 Trigonometric Functions are extremely beneficial in developing mathematical reasoning in students. With the help of these well-structured resources, students will gain a simplistic approach for efficient exam preparation. The carefully placed illustrations, questions, and examples present in these solutions are sufficient to enhance the fundamental math knowledge to excel in exams. Kids can easily plan their curriculum and revision format with the help of these solutions. To practice the exercise-wise NCERT Solutions Class 11 Maths Trigonometric Functions, try the links given below.

• Class 11 Maths Chapter 3 Ex 3.1 - 7 Questions
• Class 11 Maths Chapter 3 Ex 3.2 - 10 Questions
• Class 11 Maths Chapter 3 Ex 3.3 - 25 Questions
• Class 11 Maths Chapter 3 Ex 3.4 - 9 Questions
• Class 11 Maths Chapter 3 Miscellaneous Ex - 10 Questions

☛  Download Class 11 Maths Chapter 3 NCERT Book

Topics Covered: NCERT solutions Class 11 Maths Chapter 3 Trigonometric Functions cover some important topics, including an introduction to trigonometric ratios and identities, the measure of angles, signs, domain and range of trigonometric functions . The other important topic included in these solutions is the trigonometric solutions of the sum and difference of two angles using formulas .

Total Questions: Class 11 Maths Chapter 3 Trigonometric Functions has 51 questions in 4 exercises plus 10 questions in one miscellaneous exercise. These are primarily based on the representation of trigonometric functions, their applications, and formulas.

List of Formulas in NCERT Solutions Class 11 Maths Chapter 3

Memorizing important formulas and concepts is necessary to understand the terms and operations related to trigonometric functions clearly. NCERT Solutions Class 11 Maths Chapter 3 will provide detailed knowledge of all these along with their applications through interesting illustrations. Each concept in these solutions is well-explained with suitable examples and sample problems to promote an in-depth understanding of this topic. Formulas form an integral part of this lesson hence, creating a formulas sheet can be useful for students when they need to practice this topic. Some of the important terms, formulas, and concepts related to Trigonometric Functions explained in these solutions are listed below:

• Trigonometric Functions: Trigonometric Functions are real functions that relate the angle of a r ight-angled triangle to the ratio of the length of its sides. Sin, Cos, and Tan are the three primary functions.
• Trigonometric Equations: Trigonometric equations are equations that involve the use of trigonometric functions. These trigonometric equations are also known as trigonometric identities when the values of unknown angles for which the functions are defined are satisfied.
• Trigonometric Identities: A trigonometric equation that involves the sum and difference of angles represent a trigonometric identity. For example, sin 2 θ + cos 2 θ = 1
• Sum and Difference Identities: The sum and difference identities include the following formulas.
• sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
• cos(x+y) = cos(x)cos(y) – sin(x)sin(y)
• tan(x+y) = (tan x + tan y)/ (1−tan x • tan y)
• sin(x–y) = sin(x)cos(y) – cos(x)sin(y)
• cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
• tan(x−y) = (tan x–tan y)/ (1+tan x • tan y)

FAQs on NCERT Solutions Class 11 Maths Chapter 3

What is the importance of ncert solutions for class 11 maths chapter 3 trigonometric functions.

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are designed by well-qualified teachers and math experts to promote proficient math learning. Each exercise of these solutions is formulated as per the CBSE guidelines to support in-depth learning of Trigonometric Functions and their relations. These competently structured solutions are suitable to enhance math proficiency in students. Regular practice of these solutions will boost students’ confidence in scoring well.

What are the Important Topics Covered in NCERT Solutions Class 11 Maths Chapter 3?

NCERT Solutions Class 11 Maths Chapter 3 briefly introduces trigonometric ratios and identities along with some core concepts previously studied in grade 10. The important topics covered in these solutions are angle measures, trigonometric functions, their signs, domain, and range. The sum and difference of two angles using trigonometric functions are also included in this chapter.

Do I Need to Practice all Questions Provided in Class 11 Maths NCERT Solutions Trigonometric Functions?

NCERT Solutions Class 11 Maths Trigonometric Functions are designed in an effective way to facilitate the in-depth learning of concepts. Every question included in these solutions is aimed to improve the conceptual clarity for students to master them easily. The fundamental knowledge of Trigonometric Functions and their use will allow students to apply their knowledge in practical situations. These solutions also form the basis for studying advanced math topics, including calculus. Thus, all sums must be practiced with laser focus.

How Many Questions are there in Class 11 Maths NCERT Solutions Chapter 3 Trigonometric Functions?

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions has 61 questions in 5 exercises. These problems are sufficient to provide a deep-seated understanding of the entire topic of trigonometric functions, including important formulas, identities, ratios, and operations related to Trigonometric Functions. The jargon-free and lucid math vocabulary used in these solutions are quite proficient in easily imparting a clear step-by-step understanding of each topic as well as subtopics.

What are the Important Formulas in Class 11 Maths NCERT Solutions Chapter 3?

NCERT Solutions Class 11 Maths Chapter 3 explains the formulas related to the sum and difference of two angles in trigonometric functions. Some of the important concepts related to this topic are based on deriving expressions for trigonometric identities of the sum and difference of angles. These vital concepts are described comprehensively with the help of interesting examples for students to grasp them better. Additionally, kids need to also revise the formulas they have encountered in previous classes as those are also necessary to attempt class 11 trigonometry problems.

Why Should I Practice NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions?

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions is a reliable source of learning that gives full guidance for excellent exam preparation. With the help of these solutions, students can cover the whole syllabus of CBSE Class 11 Maths Chapter 3. The format of these solutions is quite intuitive to promote simple and easy learning. Thus, to get the best results kids need to practice these solutions.

CBSE Class 11 Maths – Chapter 3 Trigonometric Functions- Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Sets : Notes and Study Materials -pdf

• Concepts of  Trigonometric Functions
• Trigonometric Functions Master File
• Trigonometric Functions Revision Notes
• R D Sharma Solution of Trigonometric Functions
• NCERT Solution  Trigonometric Functions
• NCERT  Exemplar Solution Trigonometric Functions
• Trigonometric Functions : Solved Example 1

CBSE Class 11 Maths Notes Chapter 3 Trigonometric Functions

Angle Angle is a measure of rotation of a given ray about its initial point. The original ray is called the initial side and the final position of ray after rotation is called terminal side of the angle. The point of rotation is called vertex. If the direction of rotation is anti-clockwise, the angle is said to be positive and if the direction of rotation is clockwise, then the angle is negative.

Measuring Angles There are two systems of measuring angles Sexagesimal system (degree measure): If a rotation from the initial side to terminal side is  ( 1 360 ) t h  of a revolution, the angle is said to have a measure of one degree, written as 1°. One sixtieth of a degree is called a minute, written as 1′ and one-sixtieth of a minute is called a second, written as 1″ Thus, 1° = 60′ and 1′ = 60″

Circular system (radian measure):  A radian is an angle subtended at the centre of a circle by an arc, whose length is equal to the radius of the circle. We denote 1 radian by 1°.

Relation Between Radian and Degree We know that a complete circle subtends at its centre an angle whose measure is 2π radians as well as 360°. 2π radian = 360°. Hence, π radian = 180° or 1 radian = 57° 16′ 21″ (approx) 1 degree = 0.01746 radian

Six Fundamental Trigonometric Identities

• sinx =  1 c o s e c x
• cos x =  1 s e c x
• tan x =  1 c o t x
• sin 2  x + cos 2  x = 1
• 1 + tan 2 x = sec 2  x
• 1 + cot 2  x = cosec 2  x

Trigonometric Functions – Class 11 Maths Notes

Trigonometric ratios are defined for acute angles as the ratio of the sides of a right angled triangle. The extension of trigonometric ratios to any angle in terms of radian measure (real number) are called trigonometric function. The signs of trigonometric function in different quadrants have been given in following table.

Domain and Range of Trigonometric Functions

Sine, Cosine, and Tangent of Some Angles Less Than 90°

Allied or Related Angles The angles  n π 2 ± θ  are called allied or related angle and θ ± n × (2π) are called coterminal angles. For general reduction, we have following rules, the value of trigonometric function for ( n π 2 ± θ ) is numerically equal to

• the value of the same function, if n is an even integer with the algebraic sign of the function as per the quadrant in which angle lies.
• the corresponding co-function of θ, if n is an odd integer with the algebraic sign of the function for the quadrant in which it lies, here sine and cosine, tan and cot, sec and cosec are cofunctions of each other.

Functions of Negative Angles

For any acute angle of θ. We have,

• sin(-θ) = – sinθ
• cos (-θ) = cosθ
• tan (-θ) = – tanθ
• cot (-θ) = – cotθ
• sec (-θ) = secθ
• cosec (-θ) = – cosecθ

Some Formulae Regarding Compound Angles

An angle made up of the sum or difference of two or more angles is called compound angles. The basic results in direction are called trigonometric identities as given below: (i) sin (x + y) = sin x cos y + cos x sin y (ii) sin (x – y) = sin x cos y – cos x sin y (iii) cos (x + y) = cos x cos y – sin x sin y (iv) cos (x – y) = cos x cos y + sin x sin y

(ix) sin(x + y) sin (x – y) = sin 2  x – sin 2  y = cos 2  y – cos 2  x (x) cos (x + y) cos (x – y) = cos 2  x – sin 2  y = cos 2  y – sin 2  x

Transformation Formulae

• 2 sin x cos y = sin (x + y) + sin (x – y)
• 2 cos x sin y = sin (x + y) – sin (x – y)
• 2 cos x cos y = cos (x + y) + cos (x – y)
• 2 sin x sin y = cos (x – y) – cos (x + y)
• sin x + sin y = 2 sin( x + y 2 ) cos( x − y 2 )
• sin x – sin y = 2 cos( x + y 2 ) sin( x − y 2 )
• cos x + cos y = 2 cos( x + y 2 ) cos( x − y 2 )
• cos x – cos y = -2 sin( x + y 2 ) sin( x − y 2 )

Trigonometric Ratios of Multiple Angles

Product of Trigonometric Ratios

• sin x sin (60° – x) sin (60° + x) =  1 4  sin 3x
• cos x cos (60° – x) cos (60° + x) =  1 4  cos 3x
• tan x tan (60° – x) tan (60° + x) = tan 3x
• cos 36° cos 72° =  1 4
• cos x . cos 2x . cos 2 2 x . cos 2 3 x … cos 2 n-1  =  s i n 2 n x 2 n s i n x

Sum of Trigonometric Ratio, if Angles are in A.P.

Trigonometric Equations Equation which involves trigonometric functions of unknown angles is known as the trigonometric equation.

Solution of a Trigonometric Equation A solution of a trigonometric equation is the value of the unknown angle that satisfies the equation. A trigonometric equation may have an infinite number of solutions.

Principal Solution The solutions of a trigonometric equation for which 0 ≤ x ≤ 2π are called principal solutions.

General Solutions A solution of a trigonometric equation, involving ‘n’ which gives all solution of a trigonometric equation is called the general solutions.

General Solutions of Trigonometric Equation

• sin x = 0 ⇔ x = nπ, n ∈ Z
• cos x = 0 ⇔ x = (2n + 1)  π 2  , n ∈ Z
• tan x = 0 ⇔ x = nπ, n ∈ Z
• sin x = sin y ⇔ x = nπ + (-1) n  y, n ∈ Z
• cos x = cos y ⇔ x = 2nπ ± y, n ∈ Z
• tan x = tan y ⇔ x = nπ ± y, n ∈ Z
• sin 2  x = sin 2  y ⇔ x = nπ ± y, n ∈ Z
• cos 2  x = cos 2  y ⇔ x = nπ ± y, n ∈ Z
• tan 2  x = tan 2  y ⇔ x = nπ ± y, n ∈ Z

Basic Rules of Triangle

In a triangle ABC, the angles are denoted by capital letters A, B and C and the lengths of sides of opposite to these angles are denoted by small letters a, b and c, respectively. Sine Rule s i n A a = s i n B b = s i n C c

Cosine Rule a 2  = b 2  + c 2  – 2bc cos A b 2  = c 2  + a 2  – 2ac cos B c 2  = a 2  + b 2  – 2ab cos C

Projection Rule a = b cos C + c cos B b = c cos A + a cos C c = a cos B + b cos A

Trigonometric Functions Class 11 MCQs Questions with Answers

Question 1. The value of cos² x + cos² y – 2cos x × cos y × cos (x + y) is (a) sin (x + y) (b) sin² (x + y) (c) sin³ (x + y) (d) sin 4 (x + y)

Answer: (b) sin² (x + y) Hint: cos² x + cos² y – 2cos x × cos y × cos(x + y) {since cos(x + y) = cos x × cos y – sin x × sin y } = cos² x + cos² y – 2cos x × cos y × (cos x × cos y – sin x × sin y) = cos² x + cos² y – 2cos² x × cos² y + 2cos x × cos y × sin x × sin y = cos² x + cos² y – cos² x × cos² y – cos² x × cos² y + 2cos x × cos y × sin x × sin y = (cos² x – cos² x × cos² y) + (cos² y – cos² x × cos² y) + 2cos x × cos y × sin x × sin y = cos² x(1- cos² y) + cos² y(1 – cos² x) + 2cos x × cos y × sin x × sin y = sin² y × cos² x + sin² x × cos² y + 2cos x × cos y × sin x × sin y (since sin² x + cos² x = 1 ) = sin² x × cos² y + sin² y × cos² x + 2cos x × cos y × sin x × sin y = (sin x × cos y)² + (sin y × cos x)² + 2cos x × cos y × sin x × sin y = (sin x × cos y + sin y × cos x)² = {sin (x + y)}² = sin² (x + y)

Question 2. If a×cos x + b × cos x = c, then the value of (a × sin x – b²cos x)² is (a) a² + b² + c² (b) a² – b² – c² (c) a² – b² + c² (d) a² + b² – c²

Answer: (d) a² + b² – c² Hint: We have (a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b² ⇒ c² + (a × sin x – b × cos x)² = a² + b² ⇒ (a × sin x – b × cos x)² = a² + b² – c²

Question 3. If cos a + 2cos b + cos c = 2 then a, b, c are in (a) 2b = a + c (b) b² = a × c (c) a = b = c (d) None of these

Answer: (a) 2b = a + c Hint: Given, cos A + 2 cos B + cos C = 2 ⇒ cos A + cos C = 2(1 – cos B) ⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2) ⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2) ⇒ cos((A-C)/2) = 2sin (B/2) ⇒ cos((A-C)/2) = 2cos((A+C)/2) ⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2) ⇒ 2sin(A/2)sin(C/2) = sin(B/2) ⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac ⇒ 2(s – b) = b ⇒ a + b + c – 2b = b ⇒ a + c – b = b ⇒ a + c = 2b

Question 4. The value of cos 5π is (a) 0 (b) 1 (c) -1 (d) None of these

Answer: (c) -1 Hint: Given, cos 5π = cos (π + 4π) = cos π = -1

Question 5. In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals (a) none of these (b) c/a (c) 1 (d) a/c

Answer: (c) 1 Hint: Given cosec A (sin B cos C + cos B sin C) = cosec A × sin(B+C) = cosec A × sin(180 – A) = cosec A × sin A = cosec A × 1/cosec A = 1

Question 6. If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is (a) 4 : (√5 – 1) (b) 5 : 4 (c) (√5 – 1) : 4 (d) none of these

Answer: (a) 4 : (√5 – 1) Hint: Given, the angles of a triangle be in the ratio 1 : 4 : 5 ⇒ x + 4x + 5x = 180 ⇒ 10x = 180 ⇒ x = 180/10 ⇒ x = 18 So, the angle are: 18, 72, 90 Since a : b : c = sin A : sin B : sin C ⇒ a : b : c = sin 18 : sin 72 : sin 90 ⇒ a : b : c = (√5 – 1)/4 : {√(10 + 2√5)}/4 : 1 ⇒ a : b : c = (√5 – 1) : {√(10 + 2√5)} : 4 Now, c /a = 4/(√5 – 1) ⇒ c : a = 4 : (√5 – 1)

Question 7. The value of cos 180° is (a) 0 (b) 1 (c) -1 (d) infinite

Answer: (c) -1 Hint: 180 is a standard degree generally we all know their values but if we want to go theoretically then cos(90 + x) = – sin(x) So, cos 180 = cos(90 + 90) = -sin 90 = -1 {sin 90 = 1} So, cos 180 = -1

Question 8. The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is (a) 30° (b) 90° (c) 60° (d) 120°

Answer: (b) 90° Hint: Let the lengths of the sides if ∆ABC be a, b and c Perimeter of the triangle = 2s = a + b + c = 6(sinA + sinB + sinC)/3 ⇒ (sinA + sinB + sinC) = ( a + b + c)/2 ⇒ (sinA + sinB + sinC)/( a + b + c) = 1/2 From sin formula,Using sinA/a = sinB/b = sinC/c = (sinA + sinB + sinC)/(a + b + c) = 1/2 Now, sinB/b = 1/2 Given b = 2 So, sinB/2 = 1/2 ⇒ sinB = 1 ⇒ B = π/2

Question 9: If 3 × tan(x – 15) = tan(x + 15), then the value of x is (a) 30 (b) 45 (c) 60 (d) 90

Answer: (b) 45 Hint: Given, 3×tan(x – 15) = tan(x + 15) ⇒ tan(x + 15)/tan(x – 15) = 3/1 ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = (3 + 1)/(3 – 1) ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 4/2 ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 2 ⇒ sin(x + 15 + x – 15)/sin(x + 15 – x + 15) = 2 ⇒ sin 2x/sin 30 = 2 ⇒ sin 2x/(1/2) = 2 ⇒ 2 × sin 2x = 2 ⇒ sin 2x = 1 ⇒ sin 2x = sin 90 ⇒ 2x = 90 ⇒ x = 45

Question 10. If the sides of a triangle are 13, 7, 8 the greatest angle of the triangle is (a) π/3 (b) π/2 (c) 2π/3 (d) 3π/2

Answer: (c) 2π/3 Hint: Given, the sides of a triangle are 13, 7, 8 Since greatest side has greatest angle, Now Cos A = (b² + c² – a²)/2bc ⇒ Cos A = (7² + 8² – 13²)/(2×7×8) ⇒ Cos A = (49 + 64 – 169)/(2×7×8) ⇒ Cos A = (113 – 169)/(2×7×8) ⇒ Cos A = -56/(2×56) ⇒ Cos A = -1/2 ⇒ Cos A = Cos 2π/3 ⇒ A = 2π/3 So, the greatest angle is = 2π/3

Question 11. The value of tan 20 × tan 40 × tan 80 is (a) tan 30 (b) tan 60 (c) 2 tan 30 (d) 2 tan 60

Answer: (b) tan 60 Hint: Given, tan 20 × tan 40 × tan 80 = tan 40 × tan 80 × tan 20 = [{sin 40 × sin 80}/{cos 40 × cos 80}] × (sin 20/cos 20) = [{2 * sin 40 × sin 80}/{2 × cos 40 × cos 80}] × (sin 20/cos 20) = [{cos 40 – cos 120}/{cos 120 + cos 40}] × (sin 20/cos 20) = [{cos 40 – cos (90 + 30)}/{cos (90 + 30) + cos 40}] × (sin 20/cos 20) = [{cos 40 + sin30}/{-sin30 + cos 40}] × (sin 20/cos 20) = [{(2 × cos 40 + 1)/2}/{(-1 + cos 40)/2}] × (sin 20/cos 20) = [{2 × cos 40 + 1}/{-1 + cos 40}] × (sin 20/cos 20) = [{2 × cos 40 × sin 20 + sin 20}/{-cos 20 + cos 40 × cos 20}] = (sin 60 – sin 20 + sin 20)/(-cos 20 + cos 60 + cos 20) = sin 60/cos 60 = tan 60 So, tan 20 × tan 40 × tan 80 = tan 60

Question 12. If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is (a) 4 : (√5 – 1) (b) 5 : 4 (c) (√5 – 1) : 4 (d) none of these

Question 13. The general solution of √3 cos x – sin x = 1 is (a) x = n × π + (-1)n × (π/6) (b) x = π/3 – n × π + (-1)n × (π/6) (c) x = π/3 + n × π + (-1)n × (π/6) (d) x = π/3 – n × π + (π/6)

Answer: (c) x = π/3 + n × π + (-1)n × (π/6) Hint: √3 cos x-sin x=1 ⇒ (√3/2)cos x – (1/2)sin x = 1/2 ⇒ sin 60 × cos x – cos 60 × sin x = 1/2 ⇒ sin (x – 60) = 1/2 ⇒ sin (x – π/3) = sin 30 ⇒ sin (x – π/3) = sinπ/6 ⇒ x – π/3 = n × π + (-1)n × (π/6) {where n ∈ Z} ⇒ x = π/3 + n × π + (-1)n × (π/6)

Question 14. If tan² θ = 1 – e², then the value of sec θ + tan³ θ × cosec θ is (a) 2 – e² (b) (2 – e²) 1/2 (c) (2 – e²)² (d) (2 – e²) 3/2

Question 15. The value of cos 20 + 2sin² 55 – √2 sin65 is (a) 0 (b) 1 (c) -1 (d) None of these

Answer: (b) 1 Hint: Given, cos 20 + 2sin² 55 – √2 sin65 = cos 20 + 1 – cos 110 – √2 sin65 {since cos 2x = 1 – 2sin² x} = 1 + cos 20 – cos 110 – √2 sin65 = 1 – 2 × sin {(20 + 110)/2 × sin{(20 – 110)/2} – √2 sin65 {Apply cos C – cos D formula} = 1 – 2 × sin 65 × sin (-45) – √2 sin65 = 1 + 2 × sin 65 × sin 45 – √2 sin65 = 1 + (2 × sin 65)/√2 – √2 sin65 = 1 + √2 ( sin 65 – √2 sin 65 = 1 So, cos 20 + 2sin² 55 – √2 sin65 = 1

Question 16. If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is (a) 2π/3 (b) π/3 (c) π/2 (d) π/6

Answer: (a) 2π/3 Hint: Let S be the center of the circumcircle of triangle PQR. So, SP = SQ = SR = PQ = PR, where SP, SQ & SR are radii. Thus SPQ & SPR are equilateral triangles. ⇒ ∠QSP = 60°; Similarly ∠RQP = 60° ⇒ Angle at the center QSP = 120° So, SRPQ is a rhombus, since all the four sides are equal. Hence, its opposite angles are equal; so ∠P = ∠QSP = 120°

Question 17. If cos a + 2cos b + cos c = 2 then a, b, c are in (a) 2b = a + c (b) b² = a × c (c) a = b = c (d) None of these

Answer: (a) 2b = a + c Hint: Given, cos A + 2 cos B + cos C = 2 ⇒ cos A + cos C = 2(1 – cos B) ⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin² (B/2) ⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2) ⇒ cos((A-C)/2) = 2sin (B/2) ⇒ cos((A-C)/2) = 2cos((A+C)/2) ⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2) ⇒ 2sin(A/2)sin(C/2) = sin(B/2) ⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac ⇒ 2(s – b) = b ⇒ a + b + c – 2b = b ⇒ a + c – b = b ⇒ a + c = 2b

Question 18. The value of 4 × sin x × sin(x + π/3) × sin(x + 2π/3) is (a) sin x (b) sin 2x (c) sin 3x (d) sin 4x

Answer: (c) sin 3x Hint: Given, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = 4 × sin x × {sin x × cos π/3 + cos x × sin π/3} × {sin x × cos 2π/3 + cos x × sin 2π/3} = 4 × sin x × {(sin x)/2 + (√3 × cos x)/2} × {-(sin x)/2 + (√3 × cos x)/2} = 4 × sin x × {-(sin 2x)/4 + (3 × cos 2x)/4} = sin x × {-sin 2x + 3 × cos 2x} = sin x × {-sin 2x + 3 × (1 – sin 2x)} = sin x × {-sin 2x + 3 – 3 × sin 2x} = sin x × {3 – 4 × sin 2x} = 3 × sin x – 4 sin 3x = sin 3x So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x

Question 19. If tan A – tan B = x and cot B – cot A = y, then the value of cot (A – B) is (a) x + y (b) 1/x + y (c) x + 1/y (d) 1/x + 1/y

Answer: (d) 1/x + 1/y Hint: Given, tan A – tan B = x ……………. 1 and cot B – cot A = y ……………. 2 From equation, 1/cot A – 1/cot B = x ⇒ (cot B – cot A)/(cot A × cot B) = x ⇒ y/(cot A × cot B) = x {from equation 2} ⇒ y = x × (cot A × cot B) ⇒ cot A × cot B = y/x Now, cot (A – B) = (cot A × cot B + 1)/(cot B – cot A) ⇒ cot (A – B) = (y/x + 1)/y ⇒ cot (A – B) = (y/x) × (1/y) + 1/y ⇒ cot (A – B) = 1/x + 1/y

Question 20. The value of (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) is (a) tan 6x (b) 2 tan 6x (c) 3 tan 6x (d) 4 tan 6x

Answer: (b) 2 tan 6x Hint: Given, (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) ⇒ [{2 × sin(7x+5x)/2 × cos(7x-5x)/2}/{2 × cos(7x+5x)/2 × cos(7x-5x)/2}] + [{2 × sin(9x+3x)/2 × cos(9x-3x)/2}/{2 × cos(9x+3x)/2 × cos(9x-3x)/2}] ⇒ [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] + [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] ⇒ (sin 6x/cos 6x) + (sin 6x/cos 6x) ⇒ tan 6x + tan 6x ⇒ 2 tan 6x

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September 22, 2019 by phani

Get Free NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1, Ex 3.2, Ex 3.3, Ex 3.4, and Miscellaneous Exercise PDF in Hindi and English Medium. Trigonometric Functions Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. Trigonometric Functions All Exercises Class 11 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers.

Class 11 Maths Trigonometric Functions NCERT Solutions in English Medium and Hindi Medium

• Trigonometric Functions Class 11 Ex 3.1
• Trigonometric Functions Class 11 Ex 3.2
• Trigonometric Functions Class 11 Ex 3.3
• Trigonometric Functions Class 11 Ex 3.4
• Trigonometric Functions Class 11 Miscellaneous Exercise

त्रिकोणमितीय फलन प्रश्नावली 3.1 का हल हिंदी में

• त्रिकोणमितीय फलन प्रश्नावली 3.2 का हल हिंदी में
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• त्रिकोणमितीय फलन विविध प्रश्नावली का हल हिंदी में
• Trigonometry Formulas
• Trigonometry Functions Class 11 Notes
• NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions
• Trig Cheat Sheet
• JEE Main Trigonometry Previous Year Questions

Free download NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1, Ex 3.2, Ex 3.3, Ex 3.4, and Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.1

Ex 3.1 Class 11 Maths Question 1: Find the radian measures corresponding to the following degree measures: (i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520° Ans:

(i) \(\frac{11}{16}\) We know that: π radian = 180° ∴ \(\frac{11}{16}\) radain = \(\frac{180}{\pi} \times \frac{11}{16}\) × degree

= \(\frac{45 \times 11}{\pi \times 4}\) degree

= \(\frac{45 \times 11 \times 7}{22 \times 4}\) degree

= \(\frac{315}{8}\) degree

= 39 \(\frac{3}{8}\) degree

= 39° + \(\frac{3 \times 60}{8}\) minutes [1° = 60′]

= 39° + 22′ + \(\frac{1}{2}\) minutes

= 39°22’30” [1′ = 60°].

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Ex 3.1 Class 11 Maths Question 3: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? Ans: Number of revolutions made by the wheel in 1 minute = 360 ∴ Number of revolutions made by the wheel in 1 second = \(\frac{360}{6}\) = 6 In one complete revolution, the wheel turns an angle of 2π radian. Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e., 12π radian Thus, in one second, the wheel turns an angle of 12π radian. Ex 3.1 Class 11 Maths Question 4: Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm y an arc of length 22 cm (Use π = \(\frac{22}{7}\)). Ans:

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = \(\frac{l}{r}\) Therefore, for r = 100 cm, l = 22 cm, we have

Thus, the required angle is 12°36′.

Ex 3.1 Class 11 Maths Question 5: In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord. Ans:

Given, diameter = 40 cm ∴ radius (r) = \(\frac{40}{2}\) = 20 cm and length of chord, AB = 20 cm Thus, ∆OAB is an equilateral triangle. We know that, θ = \(\frac{\text { Arc } A B}{\text { radius }}\) ⇒ Arc AB = θ × r = \(\frac{\pi}{3}\) × 20 . = \(\frac{20}{3}\) π cm.

Ex 3.1 Class 11 Maths Question 6: If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii. Ans:

Let the radii of the two circles be r 1  and r 2 . Let an arc of length l subtend an angle of 60° at the centre of the circle of radius r 1 , while let an arc of length l subtend an angle of 75° at the centre of the circle of radius r 2 . Now, 6o° = \(\frac{\pi}{3}\) radian and 75° = \(\frac{5 \pi}{12}\) radian We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = \(\frac{l}{r}\) or l = rθ ∴ l = \(\frac{r_{1} \pi}{3}\) and

l = \(\frac{r_{2} 5 \pi}{12}\)

⇒ \(\frac{r_{1} \pi}{3}=\frac{r_{2} 5 \pi}{12}\)

⇒ r = \(\frac{r_{2} 5}{4}\)

\(\frac{r_{1}}{r_{2}}=\frac{5}{4}\) Thus, the ratio of the radii is 5 : 4.

Ex 3.1 Class 11 Maths Question 7: Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm. Ans:

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = \(\frac{l}{r}\). It is given that r = 75 cm

(i) Here, l = 10 cm θ = \(\frac{10}{75}\) radian = \(\frac{2}{15}\) radian

(ii) Here, l = 15 cm θ = \(\frac{15}{75}\) radian θ = \(\frac{1}{5}\) radian

(iii) Here, l = 21 cm θ = \(\frac{21}{75}\) radian = \(\frac{7}{75}\) radian.

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.2

sin x = \(\frac{3}{5}\)

cosec x = \(\frac{1}{\sin x}=\frac{1}{\left(\frac{3}{5}\right)}=\frac{5}{3}\) sin 2  x + cos 2  x = 1 ⇒ cos 2  x = 1 – sin 2  x ⇒ cos 2  x = 1 – (\(\frac{3}{5}\)) 2

⇒ cos 2  x = 1 – \(\frac{9}{25}\)

⇒ cos 2  x = \(\frac{16}{25}\)

⇒ cos x = ± \(\frac{4}{5}\) Since x lies in the 2nd quadrant, the value of cos x will be negative

⇒ \(\frac{4}{3}=\frac{\sin x}{\frac{-3}{5}}\) ⇒ sin x = \(\left(\frac{4}{3}\right) \times\left(\frac{-3}{5}\right)=-\frac{4}{5}\) ⇒ cosec x = \(\frac{1}{\sin x}=-\frac{5}{4}\).

tan x = – \(\frac{5}{12}\)

cot x = \(\frac{1}{\tan x}=\frac{1}{\left(-\frac{5}{12}\right)}=-\frac{12}{5}\)

Ex 3.2 Class 11 Maths Question 6: Find the value of the trigonometric function sin 765°. Ans: It is known that the values of sin x repeat after an interval of 2π or 360°. ∴ sin 765° = sin (2 × 360° + 45°) = sin 45° = 1 Ex 3.2 Class 11 Maths Question 7: Find the value of the trigonometric function cosec (- 1410°) Ans: It is known that the values of cosec x repeat after an interval of 2π or 360°. ∴ cosec (- 1410°) = cosec (- 1410° + 4 x 360°) = cosec (- 1410° + 1440°) = cosec 30° = 2. Ex 3.2 Class 11 Maths Question 8: Find the value of the trigonometric function tan \(\frac{19 \pi}{3}\). Ans:

It is known that the values of tan x repeat after an interval of π or 180°. ∴ \(\tan \frac{19 \pi}{3}=\tan 6 \frac{1}{3} \pi\)

= \(\tan \left(6 \pi+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}\)

= tan 60° = √3.

Ex 3.2 Class 11 Maths Question 9: Find the value of the trigonometric function sin \(\left(-\frac{11 \pi}{3}\right)\). Ans:

It is known that the values of cot x repeat after an interval of π or 180°.

∴ \(\sin \left(\frac{11 \pi}{3}\right)=\sin \left(-\frac{11 \pi}{3}+2 \times 2 \pi\right)\)

= \(\sin \left(\frac{\pi}{3}\right)=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\)

Ex 3.2 Class 11 Maths Question 10: Find the value of the trigonometric function cot \(\left(-\frac{15 \pi}{4}\right)\). Ans:

It is known that the values of cot x repeat after an interval of ir or 1800. ∴ \(\cot \left(-\frac{15 \pi}{4}\right)=\cot \left(-\frac{15 \pi}{4}+4 \pi\right)=\cot \frac{\pi}{4}\) = 1.

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.3

Ex 3.3 Class 11 Maths Question 1:

Prove that: sin 2  \(\frac{\pi}{6}\) + cos 2  \(\frac{\pi}{3}\) – tan 2  \(\frac{\pi}{4}\) = – \(\frac{1}{2}\) Ans:

L.H.S.= sin 2  \(\frac{\pi}{6}\) + cos 2  \(\frac{\pi}{3}\) – tan 2  \(\frac{\pi}{4}\)

= \(\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\) – (1) 2

= \(\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}\)

= R.H.S. Hence proved.

L.H.S = \(2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}\)

= \(2\left\{\sin \left(\pi-\frac{\pi}{4}\right)\right\}^{2}+2\left(\frac{1}{\sqrt{2}}\right)^{2}+2(2)^{2}\)

= \(2\left\{\sin \frac{\pi}{4}\right\}^{2}+2 \times \frac{1}{2}+8\)

= 2 \(\left(\frac{1}{\sqrt{2}}\right)^{2}\) + 1 + 8

= 1 + 1 + 8 = 10 = R.H.S. Hence proved.

Ex 3.3 Class 11 Maths Question 5: Find the value of: (i) sin 75°, (ii) tan 15° Ans:

(i) sin 75° sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° [∵ sin (x + y) = sin x cos y + cos x sin y] = \(\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\)

= \(\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

(ii) tan 15° = tan (45° – 30°)

L.H.S = \(\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}\)

= \(\frac{[-\cos x][\cos x]}{(\sin x)(-\sin x)}=\frac{-\cos ^{2} x}{-\sin ^{2} x}\)

= R.H.S Hence proved.

It is known that cos A – cos B = \(-2 \sin \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right)\)

∴ L.H.S.= \(=\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)\)

= \(– 2 \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)+\left(\frac{3 \pi}{4}-x\right)}{2}\right\} \cdot \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)-\left(\frac{3 \pi}{4}-x\right)}{2}\right\}\)

= – 2 sin (\(\frac{3 \pi}{4}\)) sin x

= – 2 sin (- \(\frac{\pi}{4}\)) sin x

= – √2 sin x = R.H.S. Hence proved.

Ex 3.3 Class 11 Maths Question 12: Prove that: sin 2  6x – sin 2  4x = sin 2x sin 10 x Ans:

It is known that sin A + sin B = 2 \(\sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

sin A – sin B = 2 \(\cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\) L.H.S.= sin 2  6x – sin 2  4x = (sin 6x + sin 4x) (sin 6x – sin 4x) = (2 sin 5x cos x) (2 cos 5x sin x) = (2 sin 5x cos 5x) (2 sin x cos x) = sin 10x sin 2x = R.H.S. Hence proved.

Ex 3.3 Class 11 Maths Question 13: Prove that: cos 2  2x cos 2  6x = sin 4x sin 8x Ans:

It is known that cos A + cos B = 2 \(\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

cos A – cos = 2 \(\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\)

∴ L.H.S = cos 2  2x – cos 2  6x = (cos 2x + cos 6x) (cos 2x – 6x) = \(\left[2 \cos \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]\left[-2 \sin \left(\frac{2 x+6 x}{2}\right) \sin \frac{(2 x-6 x)}{2}\right]\) ∴ L.H.S.= cos 2  2x – cos 2  6x = (cos 2x + cos 6x) (cos 2x – 6x) = [2 cos 4x cos (-2x)] [- 2 sin 4x sin (- 2x)] = [2 cos 4x cos 2x] [- 2 sin 4x (- sin 2x)] = (2 sin 4x cos 4x) (2 sin 2x cos 2x) = sin 8x sin 4x = R.H.S. Hence proved.

It is known that sin A + sin = 2 \(\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

cos A + cos = 2 \(\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

∴ L.H.S = \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\)

= \(\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}\)

= \(\frac{2 \sin 4 x \cos x}{2 \cos 4 x \cos x}=\frac{\sin 4 x}{\cos 4 x}\)

= tan 4x = R.H.S. Hence proved.

It is known that sin A – sin B = 2 \(\cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\)

cos 2  A – sin 2  A = cos 2A

∴ L.H.S. = \(=\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}\)

= \(\frac{2 \cos \left(\frac{x+3 x}{2}\right) \sin \left(\frac{x-3 x}{2}\right)}{-\cos 2 x}\)

= \(\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}\)

= – 2 × (- sin x) = 2 sin x

Hence proved.

Class 11 Maths NCERT Solutions Chapter 3 Exercise 3.4

cos 4x = cos 2x cos 4x – cos 2x = 0 – 2 sin \(\left(\frac{4 x+2 x}{2}\right)\) sin \(\left(\frac{4 x-2 x}{2}\right)\) = 0

[∵ cos A – cos B = 2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\(\)]

sin 3x sin x = 0 sin 3x = 0or sin x = 0 3x = nπ or x = nπ, where n ∈ Z x = \(\frac{n \pi}{3}\) or x = nπ, where n ∈ Z.

Ex 3.4 Class 11 Maths Question 6: Find the general solution of the equation cos 3x + cosx – cos 2x = 0 Ans:

cos 3x + cos x – cos 2x = 0 2 cos \(\left(\frac{3 x+x}{2}\right)\) cos \(\left(\frac{3 x-x}{2}\right)\) – cos 2x = 0

[∵ cos A + cos B = 2 \(\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)]

2 cos 2x cos x – cos 2x = 0 cos 2x (2 cos x – 1) = 0 cos 2x = 0 or 2 cos x – 1 = 0 cos 2x = 0 or cos x = \(\frac{1{2}\) ∴ 2x = (2n + 1) \(\frac{\pi}{2}\) or cos x = cos \(\frac{\pi}{3}\), where n ∈ Z x = (2n + 1) \(\frac{\pi}{4}\) or x = 2nπ ± \(\frac{\pi}{3}\) where n ∈ Z.

Ex 3.4 Class 11 Maths Question 7: Find the general solution of the equation sin 2x + cos x = 0 Ans:

sin 2x + cos x = 0 ⇒ 2sin x cos x + cos x = 0 ⇒ cos x (2 sin x + 1) = 0 ⇒ cos x = 0 or 2 sin x + 1 = 0 Now, cos x = 0 ⇒ x = (2n + 1) \(\frac{\pi}{2}\) , where n ∈ Z. or 2 sin x + 1 = 0 ⇒ sin x = – \(\frac{1}{2}\)

= – sin \(\frac{\pi}{6}\)

= sin (π + \(\frac{\pi}{6}\))

= sin \(\frac{7 \pi}{6}\)

x = nπ + (- 1) n  \(\frac{7 \pi}{6}\) where n ∈ Z Therefore, the general solution is (2n + 1) \(\frac{\pi}{2}\) or nπ + (- 1) n  \(\frac{7 \pi}{6}\) where n ∈ Z.

Ex 3.4 Class 11 Maths Question 8: Find the general solution of the equation sec 2  2x = 1 – tan 2x. Ans:

sec 2  2x = 1 – tan 2x 1 + tan 2  2x = 1 – tan 2x tan 2  x + tan 2x = 0 => tan 2x (tan 2x + 1) = 0 => tan 2x = 0 or tan 2x + 1 = 0 Now, tan 2x = 0 => tan 2x = tan 0 2x = nπ + 0, where n ∈ Z x = \(\frac{n \pi}{2}\), where n ∈ Z or tan 2x + 1 = 0 = tan 2x = – 1 = – tan \(\frac{\pi}{4}\)

= tan (π – \(\frac{\pi}{4}\))

= tan \(\frac{3 \pi}{4}\)

2x = nπ + \(\frac{3 \pi}{4}\) where n ∈ Z

x = \(\frac{n \pi}{2}+\frac{3 \pi}{8}\), where n ∈ Z

Therefore, the general solution is \(\frac{n \pi}{2}\) or \(\frac{n \pi}{2}+\frac{3 \pi}{8}\) where n ∈ Z.

Ex 3.4 Class 11 Maths Question 9: Find the general solution of the equation sin x + sin 3x + sin 5x = 0 Ans:

sin x + sin 3x + sin 5x = 0 ⇒ (sin x + sin 5x) + sin 3x = 0

\(\left[2 \sin \left(\frac{x+5 x}{2}\right) \cos \left(\frac{x-5 x}{2}\right)\right]\) + sin 3x = 0

[∵ sin A + sin B = 2 sin \(\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)]

2 sin 3x cos (2x) + sin 3x = 0 2 sin 3x cos 2x + sin 3x = 0 sin 3x (2 cos 2x +1) = 0 sin 3x = 0 or 2 cos 2x + 1 = 0 Now sin 3x = 0 ⇒ 3x = nπ, where n ∈ Z i.e., x = \(\frac{n \pi}{3}\) where n ∈ Z or 2 cos 2x + 1 = 0 cos 2x = \(-\frac{1}{2}\)

= – cos \(\frac{\pi}{3}\)

= cos (π – \(\frac{\pi}{3}\))

cos 2x = cos \(\frac{2 \pi}{3}\)

⇒ 2x = 2nπ ± \(\frac{2\pi}{3}\), where n ∈ Z

⇒ x = nπ ± \(\frac{\pi}{3}\), where n ∈ Z

Therefore, the general solution is \(\frac{n \pi}{3}\) or nπ ± \(\frac{\pi}{3}\), where n ∈ Z.

NCERT Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercise

L.H.S. = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = sin 3x sin x + sin 2  x + cos 3x cos x – cos 2  x = cos 3x cos x + sin 3x sin x – (cos 2  x – sin 2  x) = cos (3x – x) – cos 2x [∵ cos(A – B) = cos A cos B + sin A sin B] = cos 2x – cos 2x = 0 =R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 3:

Prove that: (cos x + cos y) 2  + (sin x – sin y) 2  = 4 cos 2  \(\left(\frac{x+y}{2}\right)\)

L.H.S.= (cos x + cos y) 2  + (sin x – sin y) 2 = cos 2  x + cos 2  y + 2 cos x cos y + sin 2  x + sin 2  y – 2 sin x sin y = (cos 2  x + sin 2  x) + (cos 2  y + sin 2  y) + 2 (cos x cos y – sin x sin y) = 1 + 1 + 2 cos (x + y) [∵ cos (A + B) = (cos A cos B – sin A sin B)] = 2 + 2 cos (x + y) = 2 [1 + cos (x + y)] = 2[1 + \(2 \cos ^{2}\left(\frac{x+y}{2}\right)\) – 1] [∵ cos 2A = 2 cos 2  A – 1] = 4 c0s 2  \(\left(\frac{x+y}{2}\right)\) = R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 4: Prove that: (cos x – cos y) 2  + (sin x – sin y) 2  = 4 sin 2  \(\frac{x-y}{2}\) Ans:

L.H.S.= (cos x – cos y) 2  + (sin x – sin y) 2 = cos 2  x + cos 2  y – 2 cos x cos y + sin 2  x + sin 2  y – 2 sin x sin y = (cos 2  x + sin 2  x) + (cos 2  y + sin 2  y) – 2 [cos x cos y + sin x sin y] = 1 + 1 – 2 [cos (x – y)] = 2 [1 – {1 – 2 sin 2  \(\left(\frac{x-y}{2}\right)\)}] [∵ cos 2A = 1 – 2 sin 2  A] = 4 sin 2  \(\left(\frac{x-y}{2}\right)\) = R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 5: Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x Ans:

It is known that sin A + sin B = 2 \(\sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)\) ∴ L.H.S. = (sin x + sin 3x) + (sin 5x + sin 7x) = (sin x + sin 5x) + (sin 3x + sin 7x) = \(2 \sin \left(\frac{x+5 x}{2}\right)\) . \(\cos \left(\frac{x-5 x}{2}\right)+2 \sin \left(\frac{3 x+7 x}{2}\right) \cos \left(\frac{3 x-7 x}{2}\right)\) = 2 sin 3x cos (- 2x) + 2 sin 5x cos (- 2x) = 2 sin 3x cos 2x + 2 sin 5x cos 2x = 2 cos 2x [sin 3x + sin 5x] = 2 cos 2x [latex]2 \sin \left(\frac{3 x+5 x}{2}\right) \cdot \cos \left(\frac{3 x-5 x}{2}\right)[/latex] = 2 cos 2x [2 sin 4x . cos (- x)] = 4 cos 2x sin 4x cos x = R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 6: Prove that: \(\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}\) = tan 6x Ans: It is known that

प्रश्न 3. एक-पहिया एक मिनट में 360° परिक्रमण करता है तो एक सेकंड में कितने रेडियन माप का कोण बनाएगा? हल: परिक्रमण में पहिया द्वारा बना कोण = 27 रेडियन 360 परिक्रमण में पहिया द्वारा बना कोण = 360 x 2π रेडियन 1 मिनट अर्थात् 60 सेकण्ड में 360 x 2π रेडियन का कोण बनता है। 1 सेकण्ट में चहिया द्वारा बना कोण = \(\frac { 360\times 2\pi }{ 60 }\) = 12π रेडियन।

Exercise 3.1

Q.1: Calculate the radian measurement of the given degree measurement:

(ii). 240 ∘

(iii). − 47 ∘ 30 ‘

(iv). 520 ∘

Q.2: Calculate the degree measurement of the given degree measurement: [Use π = \(\\ \frac { 22 }{ 7 } \)]

(i) \(\\ \frac { 11 }{ 16 } \)

(iii) \(\frac { 5\pi }{ 3 } \)

(iv) \(\frac { 7\pi }{ 6 } \)

Q.3: In a minute, wheel makes 360 revolutions. Through how many radians does it turn in 1 second?

Q.4: Calculate the degree measurement of the angle subtended at the centre of a circle of radius 100 m by an arc of length 22 m.

Q.5: In a circle of diameter 40 m, the length of the chord 20 m. Find the length of minor arc of chord.

Q.6: In two circles, arcs which has same length subtended at an angle of 60 ∘ and 75 ∘ at the center. Calculate the ratio of their radii.

Q.7: Calculate the angle in radian through which a pendulum swings if the length is 75 cm and the tip describes an arc of length

(iii) 21 cm

Exercise 3.2

Q.1: Calculate the values of five trigonometric func. if cos y = \(– \frac { 1 }{ 2 } \) and y lies in 3 rd quadrant.

(iii) cosec y

Q.2: Calculate the other five trigonometric function if we are given the values for sin y = \(\\ \frac { 3 }{ 5 } \), where y lies in second quadrant.

Q.3: Find the values of other five trigonometric functions if c o t y =\(\\ \frac { 3 }{ 4 } \) , where y lies in the third quadrant.

Q.4: Find the values of other five trigonometric if s e c y =\(\\ \frac { 13 }{ 5 } \) , where y lies in the fourth quadrant.

Q.5: Find the values of other five trigonometric function if tan y = \(– \frac { 5 }{ 12 } \) and y lies in second quadrant.

Q.6: Calculate the value of trigonometric function sin 765°.

Q.7: Calculate the value of trigonometric function cosec [-1410°]

Q.8: Calculate the value of the trigonometric function tan \(\frac { 19\pi }{ 3 } \) .

Q.9: Calculate the value of the trigonometric function sin \(-\frac { 11\pi }{ 3 } \) .

Q.10: Calculate the value of the trigonometric function cot \(-\frac { 15\pi }{ 4 } \)

Exercise 3.3

Q.1: Prove:

sin²\(\frac { \pi }{ 6 } \) + cos² \(\frac { \pi }{ 3 } \) – tan² \(\frac { \pi }{ 4 } \) = \(– \frac { 1 }{ 2 } \)

Q.2: Prove:

2 sin² \(\frac { \pi }{ 6 } \) + c o s e c ² \(\frac { 7\pi }{ 6 } \) 6 cos ² \(\frac { \pi }{ 3 } \) =\(\\ \frac { 3 }{ 2 } \)

Q.3: Prove:

cot ² \(\frac { \pi }{ 6 } \) + c o s e c \(\frac { 5\pi }{ 6 } \) + 3 tan ² latex s=2]\frac { \pi }{ 6 } [/latex] = 6

Q.4: Prove:

2 sin ² \(\frac { 3\pi }{ 4 } \) + 2 cos ² \(\frac { \pi }{ 4 } \) + 2 sec ² \(\frac { \pi }{ 3 } \) = 10

Q.5: Calculate the value of:

(i). sin 75 ∘

(ii). tan 15 ∘

cos ( \(\frac { \pi }{ 4 } \) – x ) cos ( \(\frac { \pi }{ 4 } \) – y ) – sin ( \(\frac { \pi }{ 4 } \) – x ) sin ( \(\frac { \pi }{ 4 } \) – y ) = sin ( x + y )

Q.7: Prove:

\(\frac { tan(\frac { \pi }{ 4 } +x) }{ tan(\frac { \pi }{ 4 } -x) } ={ \left( \frac { 1+tanx }{ 1-tanx } \right) }^{ 2 }\)

Q.8: Prove:

\(\frac { cos(\pi +x)cos(-x) }{ sin(\pi -x)cos\left( \frac { \pi }{ 2 } +x \right) } ={ cot }^{ 2 }x\)

Q.9: Prove:

\(cos(\frac { 3\pi }{ 2 } +x)cos(2\pi +x)[cot(\frac { 3\pi }{ 2 } -x)+cot(2\pi +x)]=1\)

Q.10: Prove:

sin ( n + 1 ) x sin ( n + 2 ) x + cos ( n + 1 ) x cos ( n + 2 ) x = cos x

Q.11 Prove:

\(cos(\frac { 3\pi }{ 4 } +x)-cos(\frac { 3\pi }{ 4 } -x)\)= − √2 sin x

Q.12: Prove:

sin² 6 x – sin ² 4 x = sin2 x sin 10 x

Q.13: Prove:

cos ² 2 x – cos ² 6 x = sin 4 x sin 8 x

Q.14:Prove:

sin 2 x + 2 sin 4 x + sin 6 x = 4 cos ² x sin 4 x

Q.15: Prove:

cot 4 x ( sin 5 x + sin 3 x ) = cot x ( sin 5 x – sin 3 x )

Q.16: Prove:

\(\frac { cos9x-cos5x }{ sin17x-sin3x } =-\frac { sin2x }{ cos10x } \)

Q.17: Prove:

\(\frac { sin5x+sin3x }{ cos5x+cos3x } =tan4x\)

Q.18: Prove:

\(\frac { sinx-siny }{ cosx+cosy } =tan\frac { x-y }{ 2 } \)

Q.19: Prove:

\(\frac { sinx+sin3x }{ cosx+cos3x } =tan2x\)

Q.20: Prove:

\(\frac { sinx-sin3x }{ { sin }^{ 2 }x-{ cos }^{ 2 }x } =2sinx\)

Q.21: Prove:

\(\frac { cos4x+cos3x+cos2x }{ sin4x+sin3x+sin2x } =cot3x\)

Q.22: Prove:

cot x cot 2 x – cot 2 x cot 3 x – cot 3 x cot x = 1

Q.23: Prove:

\(tan4x=\frac { 4tanx(1-{ tan }^{ 2 }x) }{ 1-6{ tan }^{ 2 }x+{ tan }^{ 4 }x } \)

Q.24: Prove:

cos 4 x = 1 – 8 sin² x cos² x

Q.25: Prove:

cos 6 x = 32 cos 6 x – 48 cos 4 x + 18 cos 2 x − 1

Exercise 3.4

Q.1: Find general solutions and the principle solutions of the given equation: tan x = √3

Q.2: Find general solutions and the principle solutions of the given equation: sec x = 2

Q.3: Find general solutions and the principle solutions of the given equation: cot = − √3

Q.4: Find general solutions and the principle solutions of the given equation: cosec x = -2

Q.5: Find the general solution of the given equation: cos 4x = cos 2x

Q.6: Find the general solution of the given equation: cos 3x + cos x – cos 2x = 0

Q.7: Find the general solution of the given equation: sin 2x + cos x = 0

Q.8: Find the general solution of the given equation: sec² 2 x = 1 – tan 2 x

Q.9: Find the general solution of the given equation: sin x + sin 3x + sin 5x = 0

Miscellaneous Exercise

Q.1: Prove that:

\(2cos\frac { \pi }{ 13 } cos\frac { 9\pi }{ 13 } +cos\frac { 3\pi }{ 13 } +cos\frac { 5\pi }{ 13 } =0\)

Q.2: Prove that:

( sin 3 x + sin x ) sin x + ( cos 3 x – cos x ) cos x = 0

Q-3: Prove that:

( cos x + cos y )² + ( sin x – sin y ) ² = 4 cos ²\(\\ \frac { x+y }{ 2 } \)

Q-4: Prove that:

( cos x – cos y ) ² + ( sin x – sin y ) ² = 4 sin ² \(\\ \frac { x-y }{ 2 } \)

Q-5: Prove that:

sin x + sin 3 x + sin 5 x + sin 7 x = 4 cos x cos 2 x cos 4 x

Q-6: Prove that:

\(\frac { (sin7x+sin5x)+(sin9x+sin3x) }{ (cos7x+cos5x)+(cos9x+cos3x) } =tan6x\)

Q-7: Show that: sin 3 y + sin 2 y – sin y = 4 sin y cos\(\\ \frac { y }{ 2 } \) cos\(\\ \frac { 3y }{ 2 } \)

Q-8: The value of tan y =\(– \frac { 4 }{ 2 } \) where y in in 2 nd quadrant then find out the values of sin\(\\ \frac { y }{ 2 } \) , cos\(\\ \frac { y }{ 2 } \) a n d tan\(\\ \frac { y }{ 2 } \) .

Q-9: The value of cos y = \(– \frac { 1 }{ 3 } \) where y in in 3 rd quadrant then find out the values of sin \(\\ \frac { y }{ 2 } \) , cos \(\\ \frac { y }{ 2 } \) a n d tan \(\\ \frac { y }{ 2 } \) .

Q-10: The value of sin y = \(\\ \frac { 1 }{ 4 } \) where y in in 2 nd quadrant then find out the values of sin \(\\ \frac { y }{ 2 } \) , cos \(\\ \frac { y }{ 2 } \) a n d tan \(\\ \frac { y }{ 2 } \) .

NCERT Solutions for Class 11 Maths All Chapters

• Chapter 1 Sets
• Chapter 2 Relations and Functions
• Chapter 3 Trigonometric Functions
• Chapter 4 Principle of Mathematical Induction
• Chapter 5 Complex Numbers and Quadratic Equations
• Chapter 6 Linear Inequalities
• Chapter 7 Permutation and Combinations
• Chapter 8 Binomial Theorem
• Chapter 9 Sequences and Series
• Chapter 10 Straight Lines
• Chapter 11 Conic Sections
• Chapter 12 Introduction to Three Dimensional Geometry
• Chapter 13 Limits and Derivatives
• Chapter 14 Mathematical Reasoning
• Chapter 15 Statistics
• Chapter 16 Probability

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Important Questions for Class 11 Maths Chapter 3 Trigonometric Functions - PDF Download

CBSE class 11 maths chapter 3 Trigonometric Functions important questions are prepared for the students who are preparing for class 11 maths exams. Trigonometric functions class 11th maths important questions have been developed by subject experts of eSaral for enhancing the problem-solving ability to score good marks in exams. You can go through the trigonometry questions for class 11 which are based on the updated syllabus of CBSE. These important questions are solved in a step by step method which helps you to understand the concepts easily.

Class 11 maths chapter 3 Trigonometric Functions is one of the easiest chapters which includes questions based on different formulas. To solve trigonometric questions for class 11, you have to learn and remember all the essential formulas of trigonometric functions. Each and every concept of trigonometric functions has been covered in important questions of class 11 mathematics chapter 3 curated by eSaral that will help you to achieve great marks in final exams. 

Students can refer to trigonometry questions with answers pdf provided by experts of mathematics. In class 11th trigonometry important questions, all the concepts are explained in precise language. This will give you a quick understanding of all the concepts and formulas while solving these important questions. Our expert teachers of mathematics at eSaral have also provided important questions of trigonometry class 11 in free PDF format which you can download from the official website of eSaral and practice these questions even before the main exam.  

Important Topics & Sub-topics of Trigonometric Functions Class 11 Maths

Chapter 3 Trigonometric Functions deals with the relation between angles and sides of triangles. This chapter is essential and a scoring one for the students of class 11. Thus, it is vital for students to have a thorough understanding of all the significant topics and sub-topics of trigonometric functions. In class 11 maths chapter 3, you will delve into the difficulties of trigonometric functions and concepts based on them. Our subject experts of eSaral have combined all the topics and sub-topics of trigonometric functions in the tabulated structure mentioned below.

Class 11 Maths Chapter 3 Trigonometric Functions Weightage

CBSE class 11 maths chapter 3 Trigonometric Functions forms solid foundations for advance level concepts of mathematical studies. In chapter 3, understanding the weightage of significant topics helps you to prioritize your preparations in an effective way. This chapter carries the highest marks in unit one. Thus, students should be thorough with the concepts of trigonometric functions chapter 3 class 11 maths to score desired marks in examination.

The weightage and marks distribution can vary from one education board to another one so students are advised to check the official website of CBSE to get the correct information regarding weightage of chapter 3. Chapter 3 Trigonometric Functions has topics angles, domain and range of trigonometric functions, and trigonometric functions of sum and difference of two angles which you must be well-versed to solve questions effectively and easily in exam. In order to score full marks in trigonometric functions, practice important questions of trigonometry class 11 maths provided on eSaral. You can also solve examples and questions mentioned in exercises of chapter 3 to grasp the knowledge of essential topics of trigonometric functions. 

Tips to Solve Class 11 Maths Chapter 3 Trigonometric Functions

Class 11 maths chapter 3 Trigonometric Functions is an easy chapter where you can score full marks by deeply analyzing and comprehending the main concepts of trigonometry. By solving 11th class trigonometry important questions, you will get to know some tips and tricks that help you to solve the questions quickly. Check out some useful tips provided below by eSaral’s subject experts of mathematics.

Firstly, Students should never miss the formulas of trigonometric functions as these formulas are crucial to solve the questions correctly.  

To understand the concepts of trigonometry deeply, you should use the resources like trigonometry important questions for class 11 available on eSaral.  

Trigonometric formulas are an essential part of chapter 3. Thus, forming a formula chart can be a helpful tool for students whenever they need to practice trigonometry based questions. 

There are some significant topics such as domain and range of trigonometric functions, relation between degree and radian, degree measure of angles etc. which students need to understand to solve questions without any error.

Benefits of Solving Class 11 Maths Chapter 3 Important Questions with Answers

Downloading and practicing important questions class 11 maths chapter 3 will give you an in-depth understanding of trigonometric functions. Students must solve trigonometry problems for class 11 by eSaral for better preparation of exams. Here, our subject experts of mathematics have provided numerous benefits of solving class 11 maths chapter 3 important questions which can be checked below.

Conceptual Understanding - By practicing class 11 maths important questions solidify the fundamental concepts of trigonometric functions. All the important questions of trigonometric functions class 11th have deeply included each question which describes the core principles of trigonometry. This process helps you to revise the concepts which you have studied in theoretical knowledge acquired in the classroom. 

Preparing for Examination - Class 11 maths chapter 3 important questions not only help you to reinforce the theoretical knowledge of trigonometric functions but also familiarize you with question format and type of questions asked in board exams. 

Boost Confidence - Solving class 11 maths chapter 3 important questions help students to recall all the vital concepts of trigonometry. You also get to learn the easy methods of solving trigonometric functions that boost your self-confidence.

Improve Time Management - By frequently practicing important questions of trigonometry class 11 maths chapter 3, You will be able to solve questions within the time allotted for question paper in the main exam. This will help you to improve your time management skills. 

Frequently Asked Questions

Answer 1. To solve trigonometry based questions, students must memorize all the formulas of trigonometric functions which help you to solve any question asked in examinations related to trigonometric functions. You should also practice questions of exercises mentioned in chapter 3. This provides deep comprehension of concepts of trigonometric functions.

Answer 2. There are a total of 3 exercises as well as one miscellaneous exercise which must be solved to score good marks in the final exam.

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MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers

November 16, 2020 by Prasanna

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Trigonometric Functions Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Question 1. The value of cos² x + cos² y – 2cos x × cos y × cos (x + y) is (a) sin (x + y) (b) sin² (x + y) (c) sin³ (x + y) (d) sin 4 (x + y)

Answer: (b) sin² (x + y) Hint: cos² x + cos² y – 2cos x × cos y × cos(x + y) {since cos(x + y) = cos x × cos y – sin x × sin y } = cos² x + cos² y – 2cos x × cos y × (cos x × cos y – sin x × sin y) = cos² x + cos² y – 2cos² x × cos² y + 2cos x × cos y × sin x × sin y = cos² x + cos² y – cos² x × cos² y – cos² x × cos² y + 2cos x × cos y × sin x × sin y = (cos² x – cos² x × cos² y) + (cos² y – cos² x × cos² y) + 2cos x × cos y × sin x × sin y = cos² x(1- cos² y) + cos² y(1 – cos² x) + 2cos x × cos y × sin x × sin y = sin² y × cos² x + sin² x × cos² y + 2cos x × cos y × sin x × sin y (since sin² x + cos² x = 1 ) = sin² x × cos² y + sin² y × cos² x + 2cos x × cos y × sin x × sin y = (sin x × cos y)² + (sin y × cos x)² + 2cos x × cos y × sin x × sin y = (sin x × cos y + sin y × cos x)² = {sin (x + y)}² = sin² (x + y)

Question 2. If a×cos x + b × cos x = c, then the value of (a × sin x – b²cos x)² is (a) a² + b² + c² (b) a² – b² – c² (c) a² – b² + c² (d) a² + b² – c²

Answer: (d) a² + b² – c² Hint: We have (a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b² ⇒ c² + (a × sin x – b × cos x)² = a² + b² ⇒ (a × sin x – b × cos x)² = a² + b² – c²

Answer: (a) 2b = a + c Hint: Given, cos A + 2 cos B + cos C = 2 ⇒ cos A + cos C = 2(1 – cos B) ⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2) ⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2) ⇒ cos((A-C)/2) = 2sin (B/2) ⇒ cos((A-C)/2) = 2cos((A+C)/2) ⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2) ⇒ 2sin(A/2)sin(C/2) = sin(B/2) ⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac ⇒ 2(s – b) = b ⇒ a + b + c – 2b = b ⇒ a + c – b = b ⇒ a + c = 2b

Answer: (c) 1 Hint: Given cosec A (sin B cos C + cos B sin C) = cosec A × sin(B+C) = cosec A × sin(180 – A) = cosec A × sin A = cosec A × 1/cosec A = 1

Question 6. If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is (a) 4 : (√5 – 1) (b) 5 : 4 (c) (√5 – 1) : 4 (d) none of these

Answer: (a) 4 : (√5 – 1) Hint: Given, the angles of a triangle be in the ratio 1 : 4 : 5 ⇒ x + 4x + 5x = 180 ⇒ 10x = 180 ⇒ x = 180/10 ⇒ x = 18 So, the angle are: 18, 72, 90 Since a : b : c = sin A : sin B : sin C ⇒ a : b : c = sin 18 : sin 72 : sin 90 ⇒ a : b : c = (√5 – 1)/4 : {√(10 + 2√5)}/4 : 1 ⇒ a : b : c = (√5 – 1) : {√(10 + 2√5)} : 4 Now, c /a = 4/(√5 – 1) ⇒ c : a = 4 : (√5 – 1)

Answer: (c) -1 Hint: 180 is a standard degree generally we all know their values but if we want to go theoretically then cos(90 + x) = – sin(x) So, cos 180 = cos(90 + 90) = -sin 90 = -1 {sin 90 = 1} So, cos 180 = -1

Question 9: If 3 × tan(x – 15) = tan(x + 15), then the value of x is (a) 30 (b) 45 (c) 60 (d) 90

Answer: (b) 45 Hint: Given, 3×tan(x – 15) = tan(x + 15) ⇒ tan(x + 15)/tan(x – 15) = 3/1 ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = (3 + 1)/(3 – 1) ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 4/2 ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 2 ⇒ sin(x + 15 + x – 15)/sin(x + 15 – x + 15) = 2 ⇒ sin 2x/sin 30 = 2 ⇒ sin 2x/(1/2) = 2 ⇒ 2 × sin 2x = 2 ⇒ sin 2x = 1 ⇒ sin 2x = sin 90 ⇒ 2x = 90 ⇒ x = 45

Answer: (c) 2π/3 Hint: Given, the sides of a triangle are 13, 7, 8 Since greatest side has greatest angle, Now Cos A = (b² + c² – a²)/2bc ⇒ Cos A = (7² + 8² – 13²)/(2×7×8) ⇒ Cos A = (49 + 64 – 169)/(2×7×8) ⇒ Cos A = (113 – 169)/(2×7×8) ⇒ Cos A = -56/(2×56) ⇒ Cos A = -1/2 ⇒ Cos A = Cos 2π/3 ⇒ A = 2π/3 So, the greatest angle is = 2π/3

Answer: (b) tan 60 Hint: Given, tan 20 × tan 40 × tan 80 = tan 40 × tan 80 × tan 20 = [{sin 40 × sin 80}/{cos 40 × cos 80}] × (sin 20/cos 20) = [{2 * sin 40 × sin 80}/{2 × cos 40 × cos 80}] × (sin 20/cos 20) = [{cos 40 – cos 120}/{cos 120 + cos 40}] × (sin 20/cos 20) = [{cos 40 – cos (90 + 30)}/{cos (90 + 30) + cos 40}] × (sin 20/cos 20) = [{cos 40 + sin30}/{-sin30 + cos 40}] × (sin 20/cos 20) = [{(2 × cos 40 + 1)/2}/{(-1 + cos 40)/2}] × (sin 20/cos 20) = [{2 × cos 40 + 1}/{-1 + cos 40}] × (sin 20/cos 20) = [{2 × cos 40 × sin 20 + sin 20}/{-cos 20 + cos 40 × cos 20}] = (sin 60 – sin 20 + sin 20)/(-cos 20 + cos 60 + cos 20) = sin 60/cos 60 = tan 60 So, tan 20 × tan 40 × tan 80 = tan 60

Question 12. If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is (a) 4 : (√5 – 1) (b) 5 : 4 (c) (√5 – 1) : 4 (d) none of these

Question 13. The general solution of √3 cos x – sin x = 1 is (a) x = n × π + (-1)n × (π/6) (b) x = π/3 – n × π + (-1)n × (π/6) (c) x = π/3 + n × π + (-1)n × (π/6) (d) x = π/3 – n × π + (π/6)

Answer: (c) x = π/3 + n × π + (-1)n × (π/6) Hint: √3 cos x-sin x=1 ⇒ (√3/2)cos x – (1/2)sin x = 1/2 ⇒ sin 60 × cos x – cos 60 × sin x = 1/2 ⇒ sin (x – 60) = 1/2 ⇒ sin (x – π/3) = sin 30 ⇒ sin (x – π/3) = sinπ/6 ⇒ x – π/3 = n × π + (-1)n × (π/6) {where n ∈ Z} ⇒ x = π/3 + n × π + (-1)n × (π/6)

Question 14. If tan² θ = 1 – e², then the value of sec θ + tan³ θ × cosec θ is (a) 2 – e² (b) (2 – e²) 1/2 (c) (2 – e²)² (d) (2 – e²) 3/2

Question 15. The value of cos 20 + 2sin² 55 – √2 sin65 is (a) 0 (b) 1 (c) -1 (d) None of these

Answer: (b) 1 Hint: Given, cos 20 + 2sin² 55 – √2 sin65 = cos 20 + 1 – cos 110 – √2 sin65 {since cos 2x = 1 – 2sin² x} = 1 + cos 20 – cos 110 – √2 sin65 = 1 – 2 × sin {(20 + 110)/2 × sin{(20 – 110)/2} – √2 sin65 {Apply cos C – cos D formula} = 1 – 2 × sin 65 × sin (-45) – √2 sin65 = 1 + 2 × sin 65 × sin 45 – √2 sin65 = 1 + (2 × sin 65)/√2 – √2 sin65 = 1 + √2 ( sin 65 – √2 sin 65 = 1 So, cos 20 + 2sin² 55 – √2 sin65 = 1

Answer: (a) 2b = a + c Hint: Given, cos A + 2 cos B + cos C = 2 ⇒ cos A + cos C = 2(1 – cos B) ⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin² (B/2) ⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2) ⇒ cos((A-C)/2) = 2sin (B/2) ⇒ cos((A-C)/2) = 2cos((A+C)/2) ⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2) ⇒ 2sin(A/2)sin(C/2) = sin(B/2) ⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac ⇒ 2(s – b) = b ⇒ a + b + c – 2b = b ⇒ a + c – b = b ⇒ a + c = 2b

Answer: (c) sin 3x Hint: Given, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = 4 × sin x × {sin x × cos π/3 + cos x × sin π/3} × {sin x × cos 2π/3 + cos x × sin 2π/3} = 4 × sin x × {(sin x)/2 + (√3 × cos x)/2} × {-(sin x)/2 + (√3 × cos x)/2} = 4 × sin x × {-(sin 2x)/4 + (3 × cos 2x)/4} = sin x × {-sin 2x + 3 × cos 2x} = sin x × {-sin 2x + 3 × (1 – sin 2x)} = sin x × {-sin 2x + 3 – 3 × sin 2x} = sin x × {3 – 4 × sin 2x} = 3 × sin x – 4 sin 3x = sin 3x So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x

Question 19. If tan A – tan B = x and cot B – cot A = y, then the value of cot (A – B) is (a) x + y (b) 1/x + y (c) x + 1/y (d) 1/x + 1/y

Answer: (d) 1/x + 1/y Hint: Given, tan A – tan B = x ……………. 1 and cot B – cot A = y ……………. 2 From equation, 1/cot A – 1/cot B = x ⇒ (cot B – cot A)/(cot A × cot B) = x ⇒ y/(cot A × cot B) = x {from equation 2} ⇒ y = x × (cot A × cot B) ⇒ cot A × cot B = y/x Now, cot (A – B) = (cot A × cot B + 1)/(cot B – cot A) ⇒ cot (A – B) = (y/x + 1)/y ⇒ cot (A – B) = (y/x) × (1/y) + 1/y ⇒ cot (A – B) = 1/x + 1/y

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Trigonometric Functions MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Syllabus for the Session 2023-24

CBSE Syllabus

Case Study Questions

Case Study on Sets      CS-2   CS-3   CS-4   CS-5

Case Study on Relations & Functions   CS-2   CS-3

Case Study on Trigonometric Functions

Case Study on Complex Numbers

Case Study on Linear Inequalities

Case Study on Permutations and Combinations

Case Study on Sequences & Series

Case Study on Straight Lines

Case Study on Conic Sections

Case Study on Statistics

Case Study on Probability

Pdf of  Case Studies

MCQs for Practice

Chapter 1 - Sets     

Chapter 2 - Relations & Functions

Chapter 3 - Trigonometric Functions

Chapter 4 - Complex Numbers & Quadratic Equations

Chapter 5 - Linear Inequalities

Chapter 6 - Permutations & Combinations

Chapter 7 - Binomial Theorem

Chapter 8 - Sequences & Series

Chapter 9 - Straight Line

Chapter 1 0 - Conic Sections

Chapter 1 1 - Introduction to Three-dimensional Geometry

Chapter 12 - Limits & D erivatives

Chapter 13 - Statistics

Chapter 14 - Probability

Answers of MCQs

Assertion & Reasoning Questions

Relations & Functions

• Trigonometric Functions

Complex Numbers

Linear Inequalities

Permutations & Combinations

Binomial Theorem

Sequences & Series

Straight Lines

Conic Sections

Introduction to Three-Dimensional Geometry

Limits & derivatives

Probability

Topic Wise Assignments of Previous Year Questions

Relations and Functions

Straight Line

Limits & Derivatives

Question Papers - DoE, Delhi

Session 2015-2016

            SA2(QP)         SA2(MS)         COMP(QP)         COMP(MS)

Session 201 6 -20 17

            SA1(QP)      SA2(QP)         SA2(MS)         COMP(QP)        COMP(MS)

Session 201 7 -201 8

            SA2(QP)          SA2(MS)         COMP(QP)         COMP(MS)

Session 201 8 -201 9

            SA2(QP)          SA2(MS)         COMP(QP)         COMP(MS)

Session 201 9 -20 20

            SA2(QP)          SA2(MS)        

Session 2020-21

    Video Explanation:  Part A     Part B Section III     Part B Section IV   Part B Section V

Session 2021-22

Question Paper 

Video Explanation:  Section A     Section B     Section C

Session 2022-23

  Sample Question Paper for Mid-Term Exam

Video Explanation: Section A     Section B    Section C     Section D     Section E

Mid-Term Exam:  Question Paper

Video Explanation: Section A     Section B     Section C     Section D    Section E

Practice Paper for Final Exam

Video Explanation: Section A     Section B     Section C     Section D     Section E

Final Exam:  Question Paper     Marking Scheme

Video Explanation: Section A     Section B     Section C     Section D     Section E

Support Material Issued by DoE, Delhi

Session 202 3 -24

 Mid-Term Exam:  Question Paper

Video Explanation:   Section A    Section B     Section C     Section D     Section E

Practice Paper 1 for Final Exam: Question Paper

Practice Paper 2 for Final Exam: Question Paper

Short Capsules (Notes to Revise Concepts)

Chapter 1 - Sets

Chapter 4 - Mathematical Induction

Chapter 5 - Complex Numbers & Quadratic Equations

Chapter 6 - Linear Inequalities

Chapter 7 - Permutations & Combinations

Chapter 8 - Binomial Theorem

Chapter 9 - Sequences & Series

Chapter 10 - Straight Line

Chapter 11 - Conic Sections

Chapter 12 - Introduction to Three-dimensional Geometry

Chapter 13 - Limits & derivatives

Chapter 14 - Mathematical Reasoning

Chapter 15 - Statistics

Chapter 16 - Probability

Some Basic Concepts to Revise:

Number System

Mensuration Results

Quadratic Equations

Solution of Polynomial Inequality - Wavy Curve Method

Probability__Revision of the Topics studied in Earlier Classes  

Result Sheets

Trigonometric Identities

• NCERT Class 11
• NCERT 11 Maths
• Chapter 3: Trigonometric Functions
• Chapter 3 Trigonometric Functions Miscellaneous Ex

NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions Miscellaneous Exercise

Chapter 3, Trigonometric Functions of Class 11 Maths, is categorised under the CBSE Syllabus for 2023-24. The miscellaneous exercise contains questions covering the entire topics present in the chapter. The last exercise of NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions is based on the following topics:

• Trigonometric Functions of Sum and Difference of Two Angles
• Trigonometric Equations

The solutions provided here are based on the latest guidelines of the syllabus prescribed by the CBSE. The NCERT Solutions for Class 11 can help the students in preparing well for the board examination.

NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions Miscellaneous Exercise

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Access other exercise solutions of Class 11 Maths Chapter 3 – Trigonometric Functions

The NCERT Class 11 Maths Solutions for Chapter 3 can be referred to in PDF format by checking the links below.

Exercise 3.1 Solutions 7 Questions

Exercise 3.2 Solutions 10 Questions

Exercise 3.3 Solutions 25 Questions

Exercise 3.4 Solutions 9 Questions

Access Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercise

Prove that:

2. (sin 3 x  + sin  x ) sin  x  + (cos 3 x  – cos  x ) cos  x  = 0

LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

By further calculation,

= sin 3x sin x + sin 2 x + cos 3x cos x – cos 2 x

Taking out the common terms,

= cos 3x cos x + sin 3x sin x – (cos 2 x – sin 2 x)

Using the formula

cos (A – B) = cos A cos B + sin A sin B

= cos (3x – x) – cos 2x

= cos 2x – cos 2x

LHS = (cos x + cos y) 2 + (sin x – sin y) 2

By expanding using the formula, we get

= cos 2 x + cos 2 y + 2 cos x cos y + sin 2 x + sin 2 y – 2 sin x sin y

Grouping the terms,

= (cos 2 x + sin 2 x) + (cos 2 y + sin 2 y) + 2 (cos x cos y – sin x sin y)

Using the formula cos (A + B) = (cos A cos B – sin A sin B)

= 1 + 1 + 2 cos (x + y)

= 2 + 2 cos (x + y)

Taking 2 as common

From the formula cos 2A = 2 cos 2 A – 1

LHS = (cos x – cos y) 2 + (sin x – sin y) 2

By expanding using the formula,

= cos 2 x + cos 2 y – 2 cos x cos y + sin 2 x + sin 2 y – 2 sin x sin y

= (cos 2 x + sin 2 x) + (cos 2 y + sin 2 y) – 2 (cos x cos y + sin x sin y)

Using the formula cos (A – B) = cos A cos B + sin A sin B

From formula cos 2A = 1 – 2 sin 2 A

5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

8. Find sin x/2, cos x/2 and tan x/2 in each of the following.

cos x = -3/5

From the formula,

9. cos x = -1/3, x in quadrant III

10. sin x = 1/4, x in quadrant II

Class 11 Maths NCERT supplementary or miscellaneous exercise solutions PDFs are provided here.

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Importance of Solving Case Study Questions for Class 11 Maths. Case study questions are an important aspect of mathematics education at the Class 11 level. These questions require students to apply their knowledge and skills to real-world scenarios, helping them develop critical thinking, problem-solving, and analytical skills.

Practice Questions For Class 11 Maths Chapter 3 Trigonometric Functions. Find the value of the below expression. Hint: Simplify the expression to. Find the general solution of the equation 5cos 2 θ + 7sin 2 θ - 6 = 0. If θ lies in the first quadrant and cos θ = 8/17, then find the value of cos (30° + θ) + cos (45° - θ) + cos (120 ...

Case Studies on the Topic Trigonometric Functions - Class 11 MathematicsThis video lecture discusses 2 case study questions on the topic of Trigonometric Fun...

Class 11 Mathematics case study question 2. A state cricket authority has to choose a team of 11 members, to do it so the authority asks 2 coaches of a government academy to select the team members that have experience as well as the best performers in last 15 matches. They can make up a team of 11 cricketers amongst 15 possible candidates.

CBSE Class 11 Maths Chapter-3 Important Questions - Free PDF Download. Trigonometric functions class 11 important questions have been prepared for students of class 11 to help them score better marks in the examination. The complete topic of trigonometric functions is designed by the subject experts following the latest guidelines of CBSE.

NCERT Solutions of Chapter 3 Class 11 Trigonometry is available free at teachoo. You can check the detailed explanation of all questions of exercises, examples and miscellaneous by clicking on the Exercise link below. We had learned Basics of Trigonometry in Class 10. In this chapter, we will learn. What is a positive or a negative angle.

4 Marks Questions. Prove the following Identities. 1.The minute hand of a watch is 1.5 cm long. How far does it tip move in 40 minute? Ans. r = 1.5 cm Angle made in 60 mint = 360 0. Angle made in 1 min = = 60 0 Angle made in 40 mint = 6 40 = 240 0. Θ =

Exercises under NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions. Exercise 3.1: In this exercise, students are introduced to trigonometric ratios of acute angles and their applications in solving problems related to heights and distances.

Students can view the set of important questions given below. Q1. Prove that sin5x−2sin3x+sinx / cos5x−cosx =tanx. A1. Starting with the left-hand side and using the trigonometric difference identities for the sine function, we obtain. L.H.S.=sin5x+sinx−2sin3x / cos5x−cosx. =2sin3x.cos2x−2sin3x / −2sin3x.sin2x.

NCERT Solutions for Class 11 Maths Chapter 3 PDF. Trigonometry sees the use of many formulas, theorems, and steps to solve the sums hence, ensuring that kids allot an ample amount of time for practice is very important. NCERT Solutions for Class 11 Maths Chapter 3 are proficiently modeled to support higher-level math learning.

Chapter 3, Trigonometric Functions, comes under NCERT Class 11 Maths and is an important chapter for students. Though the chapter has numerous mathematical terms and formulae, BYJU'S has made NCERT Solutions for Class 11 Maths easy for the students to understand and remember with the usage of tricks.

CBSE Class 11 Maths Notes Chapter 3 Trigonometric Functions. Angle. Angle is a measure of rotation of a given ray about its initial point. The original ray is called the initial side and the final position of ray after rotation is called terminal side of the angle. The point of rotation is called vertex.

Class XI Maths www.vedantu.com 4 10. Express 2cos4xsin2x as an algebraic sum of sines or cosine. Ans: x x x 11. Write the range of sT Ans: The cosine function is a periodic function with a domain of and a range of > 1@. 12. What is domain of cT Ans: The secant function is the reciprocal of the cosine function, it has a domain of (2n 1) ;n 2 ­½S

Important Questions For Class 11 Maths ... word which means 'Three', 'Gon' means 'length', and 'metry' means 'measurement'. So basically, trigonometry is a study of triangles, which has angles and lengths on its side. Trigonometry basics consist of sine, cosine and tangent functions. Trigonometry for class 11 contains ...

NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions. ... Long Answer Type Questions. Q20. If sin(θ + α) = a and sin(θ + β) = b , then prove that cos2(α - β) - 4abcos(α - β) = 1-2a 2-2b 2. Sol ...

Ex 3.2 Class 11 Maths Question 1: Find the values of other five trigonometric functions if cos x = - 1 2 x lies in third quadrant. Ans: Ex 3.2 Class 11 Maths Question 2: Find the values of other five trigonometric functions if sin x = 35, x lies in second quadrant. Ans: sin x = 3 5. cosec x = 1 sin x = 1(3 5) = 5 3.

Chapter 3 Trigonometric Functions deals with the relation between angles and sides of triangles. This chapter is essential and a scoring one for the students of class 11. Thus, it is vital for students to have a thorough understanding of all the significant topics and sub-topics of trigonometric functions. In class 11 maths chapter 3, you will ...

CLASSS 11 MATHS - CASE STUDY QUESTIONS - Free download as PDF File (.pdf), Text File (.txt) or read online for free.

We have provided Trigonometric Functions Class 11 Maths MCQs Questions with Answers to help students understand the concept very well. Trigonometric Functions Class 11 MCQs Questions with Answers. Question 1. The value of cos² x + cos² y - 2cos x × cos y × cos (x + y) is (a) sin (x + y) (b) sin² (x + y)

Class 11. Syllabus for the Session 2023-24. CBSE Syllabus. Case Study Questions. Case Study on Sets CS-2 CS-3 CS-4 CS-5. Case Study on Relations & Functions CS-2 CS-3. Case Study on Trigonometric Functions. Case Study on Complex Numbers. Case Study on Linear Inequalities.

MCQs for Chapter 3 Trigonometric Functions Class 11 with Answers. 1. If sin θ and cos θ are the roots of ax2 - bx + c = 0, then the relation between a, b and c will be. 2. If tan A = 1/2 and tan B = 1/3, then the value of A + B is. 3. The value of cos 1° cos 2° cos 3° … cos 179° is. 4.

Nov 10, 2021 • 59m • 91 views. Use Code VMSIR to Unlock this Class . In this Class , Vishal Mahajan discuss the Case Study Based Questions .This Session will be beneficial Of Class 11 & all aspirants preparing for Competitive Exams.This session will be Conducted in English & Hindi and notes will be provided in English

Solution: 10. sin x = 1/4, x in quadrant II. Solution: Class 11 Maths NCERT supplementary or miscellaneous exercise solutions PDFs are provided here. NCERT Solutions Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise can be accessed here. Students can also download these NCERT Solutions as a PDF for free.