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#### Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 1 sub question (i) maths textbook solution

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

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

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

Explanation:

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

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

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

Let,

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

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

and   $\sin (\frac{\pi }{6}) =\frac{1}{2}$

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

Hence,        \begin{aligned} &\therefore \sin ^{-1}(\sin \theta)=\theta \text { with } \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ &\sin ^{-1}\left(\sin \frac{\pi}{6}\right) \text { is } \frac{\pi}{6} \end{aligned}