$\sin^{-1}\left(\cos\frac{3\pi}{5}\right)$ equals

$\frac{-\pi}{10}$

The correct answer is Option (2) → $\frac{-\pi}{10}$

$\cos\frac{3\pi}{5} = \cos(\pi - \frac{2\pi}{5}) = -\cos\frac{2\pi}{5}$

$\sin^{-1}(\cos\frac{3\pi}{5}) = \sin^{-1}(-\cos\frac{2\pi}{5}) = -\sin^{-1}(\cos\frac{2\pi}{5})$

$\cos\frac{2\pi}{5} = \sin\frac{\pi}{2} - \frac{2\pi}{5} = \sin(\frac{\pi}{10})$

$\sin^{-1}(\cos\frac{3\pi}{5}) = -\sin^{-1}(\sin\frac{\pi}{10})$

$= -\frac{\pi}{10}$