If θ, ∈ ≤ft[(π)/2, (3π)/2], then sin^-1(sin θ ) equals

We have,

$ θ, ∈ \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]⇒ -θ, ∈\left[-\frac{3\pi}{2}, -\frac{\pi}{2}\right]⇒ \pi - θ, ∈\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

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

$∴ sin^{-1}(sin\theta )= sin^{-1} (sin(\pi - \theta )) = \pi - \theta $.