Mark the tick against the correct answer in the following:

To Find: The value of $\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$

Now, let $x=\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$

$\Rightarrow \sin x=\sin \left(\frac{2 \pi}{3}\right)$

Here range of principle value of sine is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

$\Rightarrow x=\frac{2 \pi}{3} \notin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Hence for all values of $x$ in range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the value of

$\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ is

$\Rightarrow \sin x=\sin \left(\pi-\frac{\pi}{3}\right)\left(\because \sin \left(\frac{2 \pi}{3}\right)=\sin \left(\pi-\frac{\pi}{3}\right)\right)$

$\Rightarrow \sin x=\sin \left(\frac{\pi}{3}\right)(\because \sin (\pi-\theta)=\sin \theta$ as here $\theta$ lies in II quadrant and sine is positive)

$\Rightarrow x=\frac{\pi}{3}$