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

$\theta $ $\pi -\theta $ $2\pi -\theta $ $-\pi -\theta $

$\pi -\theta $

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 $.