If $\theta \in \left[\frac{\pi }{2},\frac{3\pi }{2}\right]$, then ${\sin }^{-1}\left(\sin \theta \right)$ equals

Solution:

We have,

$\theta \in \left[\frac{\pi }{2},\frac{3\pi }{2}\right]\Rightarrow -\theta \in \left[-\frac{3\pi }{2},-\frac{\pi }{2}\right]\Rightarrow \pi -\theta \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

Also, $\sin (\pi -\theta )=\sin \theta$

$\therefore {\sin }^{-1}(\sin \theta )={\sin }^{-1}(\sin (\pi -\theta \left)\right)=\pi -\theta$