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}$