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

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

NCERT Solutions for Class 11 Maths 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 - Permutations 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

NCERT Solution Class 11 Maths of Chapter 3 All Exercises

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.

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