# NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions (Ex 3.2) Exercise 3.2

## NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions (Ex 3.2) Exercise 3.2

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## Access NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions

### Exercise 3.2

1. Find the values of the other five trigonometric functions if  $\text{cos x=-}\frac{\text{1}}{\text{2}}$ , $x$ lies in the third quadrant.

Ans:

Here given that,  $\text{cos x=-}\frac{\text{1}}{\text{2}}$

Therefore we have,

$\text{sec x=}\frac{\text{1}}{\text{cos x}}$

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\text{1}}{\text{2}} \right)}$

$\text{=-2}$

Now we know that, $\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{si}{{\text{n}}^{\text{2}}}\text{x=1-co}{{\text{s}}^{\text{2}}}\text{x}$

Substituting  $\text{cos x=-}\frac{\text{1}}{\text{2}}$ in the formula, we obtain,

$\text{si}{{\text{n}}^{\text{2}}}\text{x=1-}{{\left( \text{-}\frac{\text{1}}{\text{2}} \right)}^{\text{2}}}$

$\text{si}{{\text{n}}^{\text{2}}}\text{x=1-}\frac{\text{1}}{\text{4}}$

$\text{=}\frac{\text{3}}{\text{4}}$

$\text{sin x= }\!\!\pm\!\!\text{ }\frac{\sqrt{\text{3}}}{\text{2}}$

Since $\text{x}$ lies in the ${{\text{3}}^{\text{rd}}}$quadrant, the value of $\sin x$ will be negative.

$\text{sin x=-}\frac{\sqrt{\text{3}}}{\text{2}}$

Therefore, $\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\sqrt{\text{3}}}{\text{2}} \right)}$

$\text{=-}\frac{\text{2}}{\sqrt{\text{3}}}$

Hence ,

$\text{tan x=}\frac{\text{sin x}}{\text{cos x}}$

$\text{=}\frac{\left( \text{-}\frac{\sqrt{\text{3}}}{\text{2}} \right)}{\left( \text{-}\frac{\text{1}}{\text{2}} \right)}$

$\text{=}\sqrt{\text{3}}$

And

$\text{cot x=}\frac{\text{1}}{\text{tan x}}$

$\text{=}\frac{\text{1}}{\sqrt{\text{3}}}$

2. Find the values of other five trigonometric functions if $\text{sin x=}\frac{\text{3}}{\text{5}}$  , $\text{x}$ lies in second quadrant.

Ans:

Here given that,  $\text{sin x=}\frac{\text{3}}{\text{5}}$

Therefore we have,

$\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$=\frac{1}{\left( \frac{3}{5} \right)}$

$=\frac{5}{3}$

Now we know that , ${{\sin }^{2}}x+{{\cos }^{2}}x=1$

Therefore we have, $\text{co}{{\text{s}}^{\text{2}}}\text{x=1-si}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\sin x=\frac{3}{5}$  in the formula, we obtain,

$\text{co}{{\text{s}}^{\text{2}}}\text{x=1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}$

$\text{co}{{\text{s}}^{\text{2}}}\text{x=1-}\frac{\text{9}}{\text{25}}$

$\text{=}\frac{\text{16}}{\text{25}}$

$\text{cos x= }\!\!\pm\!\!\text{ }\frac{\text{4}}{\text{5}}$

Since $x$ lies in the ${{2}^{nd}}$ quadrant, the value of $\cos x$ will be negative.

$\text{cos x=-}\frac{\text{4}}{\text{5}}$

Therefore, $sec x=\frac{1}{\cos x}$

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\text{4}}{\text{5}} \right)}$

$\text{=-}\frac{\text{5}}{\text{4}}$

Hence ,

$\tan x=\frac{\sin x}{\cos x}$

$\text{=}\frac{\left( \frac{\text{3}}{\text{5}} \right)}{\left( \text{-}\frac{\text{4}}{\text{5}} \right)}$

$\text{=-}\frac{\text{3}}{\text{4}}$

And

$\cot x=\frac{1}{\tan x}$

$\text{=-}\frac{\text{4}}{\text{3}}$

3. Find the values of other five trigonometric functions if $\text{cot x=}\frac{\text{3}}{\text{4}}$ , $\text{x}$ lies in third quadrant.

Ans:

Here given that,  $\cot x=\frac{3}{4}$

Therefore we have,

$\tan x=\frac{1}{\cot x}$

$=\frac{1}{\left( \frac{3}{4} \right)}$

$=\frac{4}{3}$

Now we know that , $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\text{tan x=}\frac{\text{4}}{\text{3}}$  in the formula, we obtain,

${{\sec }^{2}}x=1+{{\left( \frac{4}{3} \right)}^{2}}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}\frac{\text{16}}{\text{9}}$

$=\frac{25}{9}$

$\text{sec x= }\!\!\pm\!\!\text{ }\frac{\text{5}}{\text{3}}$

Since $x$ lies in the ${{3}^{rd}}$ quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}\frac{\text{5}}{\text{3}}$

Therefore, $\cos x=\frac{1}{\sec x}$

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\text{5}}{\text{3}} \right)}$

$\text{=-}\frac{\text{3}}{\text{5}}$

Now  , $\text{tan x=}\frac{\text{sin x}}{\text{cos x}}$

Therefore, $\text{sin x=tan xcos x}$

Hence we have, $\text{sin x=}\frac{\text{4}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\frac{\text{3}}{\text{5}} \right)$

$\text{=}\left( \text{-}\frac{\text{4}}{\text{5}} \right)$

And

$\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$\text{=-}\frac{\text{5}}{\text{4}}$

4. Find the values of other five trigonometric functions if $\text{sec x=}\frac{\text{13}}{\text{5}}$  , $\text{x}$ lies in fourth quadrant.

Ans:

Here given that,  $\sec x=\frac{13}{5}$

Therefore we have,

$\cos x=\frac{1}{\sec x}$

$=\frac{1}{\left( \frac{13}{5} \right)}$

$=\frac{5}{13}$

Now we know that , $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{ta}{{\text{n}}^{\text{2}}}\text{x=se}{{\text{c}}^{\text{2}}}\text{x-1}$

Substituting  $\text{sec x=}\frac{\text{13}}{\text{5}}$  in the formula, we obtain,

$\text{ta}{{\text{n}}^{\text{2}}}\text{x=}{{\left( \frac{\text{13}}{\text{5}} \right)}^{\text{2}}}\text{-1}$

$\text{ta}{{\text{n}}^{\text{2}}}\text{x=}\frac{\text{169}}{\text{25}}\text{-1}$

$\text{=}\frac{\text{144}}{\text{25}}$

$\text{tanx= }\!\!\pm\!\!\text{ }\frac{\text{12}}{\text{5}}$

Since $x$ lies in the ${{4}^{th}}$ quadrant, the value of $\tan x$ will be negative.

$\text{tan x=-}\frac{\text{12}}{\text{5}}$

Therefore, $\text{cot x=}\frac{\text{1}}{\text{tan x}}$

$\text{=-}\frac{\text{5}}{\text{12}}$

Now  , $\text{tan x=}\frac{\text{sin x}}{\text{cos x}}$

Therefore, $\text{sin x=tan xcos x}$

Hence we have, $\text{sin x=}\frac{\text{5}}{\text{13}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\frac{\text{12}}{\text{5}} \right)$

$\text{=}\left( \text{-}\frac{\text{12}}{\text{13}} \right)$

And

$\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$\text{=-}\frac{\text{13}}{\text{12}}$

5. Find the values of other five trigonometric functions if  $\text{tan x=-}\frac{\text{5}}{\text{12}}$ , $\text{x}$ lies in second quadrant.

Ans:

Here given that,  $\text{tan x=-}\frac{\text{5}}{\text{12}}$

Therefore we have,

$\text{cot x=}\frac{\text{1}}{\text{tan x}}$

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\text{5}}{\text{12}} \right)}$

$\text{=-}\frac{\text{12}}{\text{5}}$

Now we know that , $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\text{tan x=-}\frac{\text{5}}{\text{12}}$  in the formula, we obtain,

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}{{\left( \text{-}\frac{\text{5}}{\text{12}} \right)}^{\text{2}}}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}\frac{\text{25}}{\text{144}}$

$=\frac{169}{144}$

$\text{sec x= }\!\!\pm\!\!\text{ }\frac{\text{13}}{\text{12}}$

Since $x$ lies in the ${{2}^{nd}}$ quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}\frac{\text{13}}{\text{12}}$

Therefore, $\text{cos x=}\frac{\text{1}}{\text{sec x}}$

$\text{=-}\frac{\text{12}}{\text{13}}$

Now  , $\text{tan x=}\frac{\text{sin x}}{\text{cos x}}$

Therefore, $\text{sin x=tan xcos x}$

Hence we have, $\text{sin x=}\left( \text{-}\frac{\text{5}}{\text{12}} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-}\frac{\text{12}}{\text{13}} \right)$

$=\left( \frac{5}{13} \right)$

And

$\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$\text{=}\frac{\text{13}}{\text{5}}$

6. Find the value of the trigonometric function $\text{sin76}{{\text{5}}^{\text{o}}}$ .

Ans: We know that the values of $\sin x$ repeat after an interval of $2\pi$ or ${{360}^{\circ }}$ .

Therefore we can write,

$\text{sin76}{{\text{5}}^{\text{o}}}\text{=sin}\left( \text{2 }\!\!\times\!\!\text{ 36}{{\text{0}}^{\text{o}}}\text{+4}{{\text{5}}^{\text{o}}} \right)$

$\text{=sin4}{{\text{5}}^{\text{o}}}$

$\text{=}\frac{\text{1}}{\sqrt{\text{2}}}\text{.}$

7. Find the value of the trigonometric function $\text{cosec}\left( \text{-141}{{\text{0}}^{\text{o}}} \right)$

Ans: We  know that the values of $\text{cosec x}$ repeat after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ or ${{360}^{\circ }}$ .

Therefore we can write,

$\text{cosec}\left( \text{-141}{{\text{0}}^{\text{o}}} \right)\text{=cosec}\left( \text{-141}{{\text{0}}^{\text{o}}}\text{+4 }\!\!\times\!\!\text{ 36}{{\text{0}}^{\text{o}}} \right)$

$\text{=cosec}\left( \text{-141}{{\text{0}}^{\text{o}}}\text{+144}{{\text{0}}^{\text{o}}} \right)$

$\text{=cosec3}{{\text{0}}^{\text{o}}}$

$=2$

8. Find the value of the trigonometric function  $\text{tan}\frac{\text{19 }\!\!\pi\!\!\text{ }}{\text{3}}$ .

Ans: We know that the values of $\text{tan x}$ repeat after an interval of $\text{ }\!\!\pi\!\!\text{ }$ or ${{180}^{\circ }}$.

Therefore we can write,

$\text{tan}\frac{\text{19 }\!\!\pi\!\!\text{ }}{\text{3}}\text{=tan6}\frac{\text{1}}{\text{3}}\text{ }\!\!\pi\!\!\text{ }$

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

$\text{=tan}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

$\text{=}\sqrt{\text{3}}$

9. Find the value of the trigonometric function $\text{sin}\left( \text{-}\frac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

Ans: We know that the values of $\text{sin x}$ repeat after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ or ${{360}^{\circ }}$ .

Therefore we can write,

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

$\text{=sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

$=\frac{\sqrt{3}}{2}$

10. Find the value of the trigonometric function $\text{cot}\left( \text{-}\frac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}} \right)$

Ans: We know that the values of $\text{cot x}$ repeat after an interval of $\text{ }\!\!\pi\!\!\text{ }$ or ${{180}^{\circ }}$.

Therefore we can write,

$\text{cot}\left( \text{-}\frac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}} \right)\text{=cot}\left( \text{-}\frac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+4 }\!\!\pi\!\!\text{ } \right)$

$\text{=cot}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

$=1$

### Points to Remember Before Solving Exercise 3.2

1. Trigonometric Identities

• sin2 A + cos2 A = 1

• 1 + tan2 A = sec2 A

• 1 + cot2 A = cosec2 A

2. Domain and Range of Trigonometric Functions

The input and output values of trigonometric functions, respectively, are the domain and range of trigonometric functions. The range of trigonometric functions denotes the resultant value of the trigonometric function corresponding to a given angle in the domain, whereas the domain of trigonometric functions denotes the values of angles where the trigonometric functions are given.

3. Sign of Trigonometric Functions in Different Quadrants

All functions are positive in the first quadrant, only sin and cosec are positive in the second quadrant, and only tan and cot are positive in the third quadrant, and only cos and sec are positive in the fourth quadrant.

4. Domain and Range Table for Trigonometric Functions

 Trigonometric Functions Domain Range sin θ (-∞, + ∞) [-1, +1] cos θ (-∞, +∞) [-1, +1] tan θ R - (2n + 1)π/2 (-∞, +∞) cot θ R - nπ (-∞, +∞) sec θ R - (2n + 1)π/2 (-∞, -1] U [+1, +∞) cosec θ R - nπ (-∞, -1] U [+1, +∞)

### NCERT Solution Class 11 Maths of Chapter 3 All Exercises

 Chapter 3 - Trigonometric Functions Exercises in PDF Format Exercise 3.1 7 Questions & Solutions Exercise 3.2 10 Questions & Solutions Exercise 3.3 25 Questions & Solutions Exercise 3.4 9 Questions & Solutions Miscellaneous Exercise 10 Questions & Solutions

## NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions (Ex 3.2) Exercise 3.2

Opting for the NCERT solutions for Ex 3.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 3.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 3 Exercise 3.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

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## FAQs on NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions (Ex 3.2) Exercise 3.2

1. What are the trigonometric functions?

Ans: Usually, trigonometric calculations are done using the trigonometry ratios. There are a total six trigonometric ratios out of which two (Cos and Sin) are basic ratios. The main ratios are Sin, Cos and Tan such that Tan is derived from sin and Cos. Cosec, Sec and Cot are the inverse of the three main ratios. Cosec is the inverse of Sin, Sec is the inverse of Cos and Cot is the inverse of Tan. These ratios are also called trigonometric functions.

2. Why only right-angled triangles are used in trigonometry?

Ans: Trigonometry is used to define the relations between angles and sides of a triangle. The triangles are not specifically right-angled triangles but any triangle. The reason why we use right angle triangles to derive formulas is because the sides and angles of a right-angled triangle have fixed relationships which is shown in Pythagoras theorem. According to Pythagoras theorem, the square of the hypotenuse of a right-angled triangle is equal to the sum of the square of its base and length. The other reason for using right angled triangles is the fact that every triangle can be bisected into two or more right-angled triangles. Thus, the value of any triangles can be found if the values of the right angled triangle is given.

3. What is the application of Trigonometry in real life?

Ans: Trigonometry finds its applications in various fields. We use Trigonometry to find the angle of elevation, the height of an object at a particular distance or the distance from an object of a given height. Trigonometry is used by architecture to find the inclination of a roof, the height of a building, etc. It is used by detectives to find out the height of the criminal by knowing at which angle the bullet was shot and the distance from where it was shot. Marine biologists and marine engineers also use Trigonometry for various purposes, for example to understand the size and behaviour of sea animals, to find out the minimum distance to be maintained from huge icebergs, etc. Vedantu makes Trigonometry more interesting for you by simplifying the complicated numerical and relating each of them to real-life scenarios.

4. What are the benefits of using the NCERT solutions by Vedantu?

Ans: NCERT solutions for class 11 Maths Chapter 3 is written keeping in mind the age group of the students and thus the language used is simple, well spaced and broken into steps. It includes all the important points, formulas, terms and principles in the chapter. The answers are treated systematically and kept precise, brief and self explanatory.

5. What do you mean by Radian Measure?

Ans: Radian Measure of a circle's central angle in Class 11 Maths Chapter 3  is defined as the ratio of the length of the arc subtended by that angle to the radius of the circle. The symbol rad represents a radian. There are a lot of questions that may ask you to find radian measures. You can learn more about radian measure and the questions based on its concept through Vedantu.

6. How many questions are there in Class 11 Maths Chapter 3 Exercise 3.2?

Ans: There are a total of 10 questions in Exercise 3.2 of Class 11 Maths Chapter 3 'Trigonometric Functions'. All the questions here ask to find the value of various trigonometric functions. You can clear your concepts of the chapter by practising these questions. If you find any difficulty in solving these questions, you can refer to NCERT solutions provided by the expert team of Vedantu, free of cost.

7. How will Class 11 Maths help me in Class 12 Boards?

Ans: Class 11th holds a very important place in a student's life. It not only prepares you for Class 11 but also for Class 12 and for competitive exams as well. Math Students must remember that the concepts they learn in Class 11 are going to be very useful in Class 12. Thorough learning of topics like Calculus, Coordinate Geometry, Vectors, Trigonometry will make your foundation stronger for further studies.

8. Is Class 11 Maths Chapter 3 important?

Ans: The unit 'Sets and functions', of which Chapter 3 is a part, itself holds the majority of marks weightage in exams. Therefore, Class 11 Maths Chapter 3 is a very important chapter. As you have already studied Trigonometric identities and ratios, now it's time to move to a higher level. Since this chapter has a lot of complex theories, Vedantu is here to make it easier for you.

9. How can I avail study material for Class 11 Maths?

Ans: If you are a Maths student, you must definitely know how complex Maths becomes when you reach Class 11. Therefore, to make your learning process even easier, Vedantu comes with various study guides including NCERT solutions, important questions, sample papers, etc. You can avail the study material for Class 11 Maths at the site of Vedantu or by downloading the Vedantu app. SHARE TWEET SHARE SUBSCRIBE