#### need solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (iii)

Answer:    $\frac{\pi }{6}$

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

Given:   $\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)$

Explanation:

First, we solve  $\sin \frac{5\pi }{6}$

$\frac{5 \pi}{6}=\pi-\frac{\pi}{6}$

Then,

\begin{aligned} &\sin \left(\pi-\frac{\pi}{6}\right)=\sin \left(\frac{\pi}{6}\right) \\ &\sin (\pi-\theta)=\sin (\theta) \end{aligned}

We know that the value of  $\sin \frac{\pi }{6}$  is $\frac{1}{2}$

$\sin \frac{5 \pi}{6}=\frac{1}{2}$

By substituting this value in  $\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)$

we get     $\sin ^{-1}\left(\frac{1}{2}\right)$

Let          $y=\sin ^{-1}\left(\frac{1}{2}\right)$

\begin{aligned} &\sin y=\frac{1}{2} \\ &\sin \left(\frac{\pi}{6}\right)=\frac{1}{2} \end{aligned}

The range of principal value of  $\sin ^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text { and } \sin \left(\frac{\pi}{6}\right)=\frac{1}{2}$

\begin{aligned} &\sin ^{-1}\left(\sin \frac{\pi}{6}\right)=\frac{\pi}{6} \\ &\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)=\frac{\pi}{6} \quad \because\left[\sin ^{-1}(\sin \theta)=\theta\right] \end{aligned}