NCERT Solutions for Class 12 Maths Chapter 2: Inverse Trigonometric Functions - Exercise 2.1

Exercise 2.1

1.  Find the principal value of ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ 

Ans: Let’s assume when  ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = y$, Then $\sin y = \left( { - \dfrac{1}{2}} \right) =  - \sin \left( {\dfrac{\pi }{6}} \right) = \sin \left( { - \dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ and $\sin \left( { - \dfrac{\pi }{6}} \right) = \dfrac{1}{2}$

Hence, the principal value of ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ is $ - \dfrac{\pi }{6}$.

2. Find the principal value of ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$ 

Ans: Let’s consider, ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = y.$ Then $\cos y = \dfrac{{\sqrt 3 }}{2} = \cos \left( {\dfrac{\pi }{6}} \right)$

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $\cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}$

Therefore, the principal value of ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$ is $\dfrac{\pi }{6}$.

3. Find the principal value of ${\operatorname{cosec} ^{ - 1}}(2)$

Ans: Let’s consider,${\operatorname{cosec} ^{ - 1}}(2) = y$.Then, $\operatorname{cosec} {\text{y}} = 2 = \operatorname{cosec} \left( {\dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\operatorname{cosec} ^{ - 1}}$ is$\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] - \{ 0\} $. 

Therefore, the principal value of ${\operatorname{cosec} ^{ - 1}}(2)$ is $\dfrac{\pi }{6}$.

4. Find the principal value of ${\tan ^{ - 1}}( - \sqrt 3 )$

Ans: Let’s consider ${\tan ^{ - 1}}( - \sqrt 3 ) = y$ Then, $\tan y =  - \sqrt 3  =  - \tan \dfrac{\pi }{3} = \tan \left( { - \dfrac{\pi }{3}} \right)$.

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ and$\tan \left( { - \dfrac{\pi }{3}} \right)$ is $ - \sqrt 3 $

Hence, the principal value of ${\tan ^{ - 1}}( - \sqrt 3 )$ is $ - \dfrac{\pi }{3}$.

5. Find the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$

Ans: Let’s consider, ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = y.$

Then, $\cos y =  - \dfrac{1}{2} =  - \cos \left( {\dfrac{\pi }{3}} \right) = \cos \left( {\pi  - \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{{2\pi }}{3}} \right)$.

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $\cos \left( {\dfrac{{2\pi }}{3}} \right) =  - \dfrac{1}{2}$

Therefore, the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ is $\left( {\dfrac{{2\pi }}{3}} \right)$.

6. Find the principal value of ${\tan ^{ - 1}}( - 1)$

Ans: Let’s assume that ${\tan ^{ - 1}}( - 1) = {\text{y}}$.

Then, $\tan y =  - 1 =  - \tan \left( {\dfrac{\pi }{4}} \right) = \tan \left( { - \dfrac{\pi }{4}} \right)$.

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ and$\tan \left( { - \dfrac{\pi }{4}} \right) =  - 1$

Therefore, the principal value of ${\tan ^{ - 1}}( - 1)$ is $ - \dfrac{\pi }{4}$.

7. Find the principal value of ${\sec ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right)$

Ans: Let’s consider ${\sec ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right) = y$. 

Then, $\sec y = \dfrac{2}{{\sqrt 3 }} = \sec \left( {\dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\sec ^{ - 1}}$ is $[0,\pi ] - \left\{ {\dfrac{\pi }{2}} \right\}$ and $\sec \left( {\dfrac{\pi }{6}} \right) = \dfrac{2}{{\sqrt 3 }}$

8. Find the principal value of ${\cot ^{ - 1}}(\sqrt 3 )$

Ans: Let’s consider ${\cot ^{ - 1}}(\sqrt 3 ) = y$. Then $\cot y = \sqrt 3  = \cot \left( {\dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\cot ^{ - 1}}$ is $(0,\pi )$ and $\cot \left( {\dfrac{\pi }{6}} \right) = \sqrt 3 $.

Hence, the principal value of ${\cot ^{ - 1}}(\sqrt 3 )$ is $\dfrac{\pi }{6}$.

9. Find the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$

Ans: Let’s ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) = y$.

Then $\cos y =  - \dfrac{1}{{\sqrt 2 }} =  - \cos \left( {\dfrac{\pi }{4}} \right) = \cos \left( {\pi  - \dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{{3\pi }}{4}} \right)$.

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $\cos \left( {\dfrac{{3\pi }}{4}} \right) =  - \dfrac{1}{{\sqrt 2 }}$.

Therefore, the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$ is $\dfrac{{3\pi }}{4}$.

 10. Find the principal value of ${\operatorname{cosec} ^{ - 1}}( - \sqrt 2 )$

Ans: Let’s consider, ${\operatorname{cosec} ^{ - 1}}( - \sqrt 2 ) = y$. Then, $\cos ecy =  - \sqrt 2  =  - \cos ec\left( {\dfrac{\pi }{4}} \right) = \cos ec\left( { - \dfrac{\pi }{4}} \right)$

As we know that the range of the principal value branch of ${\operatorname{cosec} ^{ - 1}}$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] - \{ 0\} $ and $\operatorname{cosec} \left( { - \dfrac{\pi }{4}} \right) =  - \sqrt 2 $

Therefore, the principal value of ${\operatorname{cosec} ^{ - 1}}( - \sqrt 2 )$ is $ - \dfrac{\pi }{4}$.

11. Find the value of ${\tan ^{ - 1}}(1) + {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$

Ans: Let’s consider ${\tan ^{ - 1}}(1) = x$. Then, $\tan x = 1 = \tan \left( {\dfrac{\pi }{4}} \right)$.

$\therefore {\tan ^{ - 1}}(1) = \dfrac{\pi }{4}$

Let’s assume,${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = y$.

Then, $\cos y =  - \dfrac{1}{2} =  - \cos \left( {\dfrac{\pi }{3}} \right) = \cos \left( {\pi  - \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{{2\pi }}{3}} \right)$.

$\therefore {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \dfrac{{2\pi }}{3}$

Let’s again assume that ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = z$.

Then, $\sin z =  - \dfrac{1}{2} =  - \sin \left( {\dfrac{\pi }{6}} \right) = \sin \left( { - \dfrac{\pi }{6}} \right)$.

$\therefore {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) =  - \dfrac{\pi }{6}$

$\therefore {\tan ^{ - 1}}(1) + {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$

$ = \dfrac{\pi }{4} + \dfrac{{2\pi }}{3} - \dfrac{\pi }{6}$

$ = \dfrac{{3\pi  + 8\pi  - 2\pi }}{{12}} = \dfrac{{9\pi }}{{12}} = \dfrac{{3\pi }}{4}$

12. Find the value of ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$

Ans: Let’s consider,${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = x$.

Then, $\cos x = \dfrac{1}{2} = \cos \left( {\dfrac{\pi }{3}} \right)$.

$\therefore {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{3}$

Let’s assume ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = y.$

Then, $\sin y = \dfrac{1}{2} = \sin \left( {\dfrac{\pi }{6}} \right)$.

$\therefore {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}$

$\therefore {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{3} + 2 \times \dfrac{\pi }{6} = \dfrac{\pi }{3} + \dfrac{\pi }{3} = \dfrac{{2\pi }}{3}$

13. If ${\sin ^{ - 1}}x = y$, then

(A) $0 \leqslant {\text{y}} \leqslant \pi $

(B) $ - \dfrac{\pi }{2} \leqslant y \leqslant \dfrac{\pi }{2}$

(C) $0 < y < \pi $

(D) $ - \dfrac{\pi }{2} < y < \dfrac{\pi }{2}$

Ans: It is given that ${\sin ^{ - 1}}x = y$. As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$. 

Therefore, $ - \dfrac{\pi }{2} \leqslant y \leqslant \dfrac{\pi }{2}$.

Hence option (B) is correct.

14. ${\tan ^{ - 1}}\sqrt 3  - {\sec ^{ - 1}}( - 2)$ is equal to

(A) $\pi $

(B) $ - \pi /3$

(C) $\pi /3$

(D) $2\pi /3$

Ans: Let’s consider, ${\tan ^{ - 1}}\sqrt 3  = x.$.

Then, $\tan x = \sqrt 3  = \tan \dfrac{\pi }{3}$

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)$. $\therefore {\tan ^{ - 1}}\sqrt 3  = \dfrac{\pi }{3}$

Let assume, ${\sec ^{ - 1}}( - 2) = y$.

Then, $\sec y =  - 2 =  - \sec \left( {\dfrac{\pi }{3}} \right) = \sec \left( {\pi  - \dfrac{\pi }{3}} \right) = \sec \dfrac{{2\pi }}{3}$.

As we know that the range of the principal value branch of sec $^1$ is $[0,\pi ] - \left\{ {\dfrac{\pi }{2}} \right\}$. $\therefore {\sec ^{ - 1}}( - 2) = \dfrac{{2\pi }}{3}$

Thus, ${\tan ^{ - 1}}(\sqrt 3 ) - {\sec ^{ - 1}}( - 2)$

$ = \dfrac{\pi }{3} - \dfrac{{2\pi }}{3} =  - \dfrac{\pi }{3}$

NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

NCERT Solution Class 12 Maths of Chapter 2 All Exercises

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