NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions 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

Conclusion

Understanding trigonometric functions is crucial for solving problems involving angles and triangles. Focus on mastering the definitions, sign conventions, domain, range, and graphical representations of sine, cosine, and tangent functions. These concepts are foundational in fields like engineering, physics, and architecture. Previous year question papers typically include 2-3 questions on trigonometric functions, highlighting their importance in exams. Practice diverse problems and use NCERT Solutions to strengthen your understanding and preparation. Clear concepts in trigonometry will not only aid in academic success but also in practical applications in various disciplines.

Class 11 Maths Chapter 3: Exercises Breakdown

CBSE Class 11 Maths Chapter 3 Other Study Materials

Chapter-Specific NCERT Solutions for Class 11 Maths

The chapter-wise NCERT Solutions for Class 11 Maths are given below. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.