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

Answer: $-\frac{\pi }{6}$

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

Given:   $\sin ^{-1}\left(\sin \frac{7 \pi}{6}\right)$

Explanation:

First, we solve $\left(\sin \frac{7 \pi}{6}\right)$

$\operatorname{Sin} \frac{7 \pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)$

As we know $\sin \left(\frac{\pi}{6}\right)=\frac{1}{2}$

$\sin \left(\pi+\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right) \quad \because \sin [\pi+\theta]=\sin (-\theta)$

$\sin ^{-1}\left(\sin \frac{7 \pi}{6}\right)=\sin ^{-1}\left(-\frac{1}{2}\right)$

Now,    $\text { let } y=\sin ^{-1}\left(-\frac{1}{2}\right)$

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

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

$\therefore \sin ^{-1}\left(\sin \frac{\pi}{6}\right)=\frac{\pi}{6} \quad\left\{\text { As } \operatorname{Sin} \frac{7 \pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)\right\}$

As          \begin{aligned} &\left[\sin ^{-1}(\sin x)\right]=x \end{aligned}