#### Need solution for RD Sharma maths class 12 chapter Inverse Trigonometric Functions exercise 3.5 question 3 sub question (iii)

$\frac{\pi}{6}$

Hints: The principal value branch of the function $\operatorname{cosec}^{-1}$is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$

The principal value branch of the function $\sin ^{-1}$is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ]$

Given:  $\sin ^{-1}\left[\cos \left\{2 \operatorname{cosec}^{-1}(-2)\right\}\right]$

Range of $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$

$\sin ^{-1}\left[\cos \left\{2 \times \frac{-\pi}{6}\right\} .\right.$                                                                                           $\operatorname{cosec}^{-1}(\sqrt{2})= \frac{-\pi}{6}$

$\sin ^{-1}\left[\cos \left\{\frac{-\pi}{3}\right\}\right]$

$\sin ^{-1}\left[\cos \left\{\frac{-\pi}{3}\right\}\right]$                                                                                                             $[\cos( -\theta) = \cos \theta]$

\begin{aligned} &\sin ^{-1}\left(\frac{1}{2}\right)\\ &\text { Let } y=\sin ^{-1}\left(\frac{1}{2}\right)\\ &\sin y=\sin \frac{\pi}{6}\\ &y=\frac{\pi}{6}\\ &\therefore \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\\ &\text { Therefore, the principal value of }\\ &\sin ^{-1}\left[\cos \left\{2 \operatorname{cosec}^{-1}(-2)\right\}\right] \text { is } \frac{\pi}{6} \end{aligned}