STD 12 - 2. inverse trigonometric function — JEE कक्षा 11 साइन्स — Question

Then $, \sin x=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\frac{-\pi}{6}\right)$

We know that the range of the principal value branch of $\sin ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ $\sin ^{-1}\left(\frac{-1}{2}\right)=\frac{\pi}{6}$

$\therefore \sin \left(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right)=\sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)$$=\sin \left(\frac{3 \pi}{6}\right)=\sin \left(\frac{\pi}{2}\right)=1$