The value of sin^-1 ≤ft[ cos ≤ft( (33π)/5 ) ] is

The correct answer is Option (4) → $\frac{-\pi}{10}$ ##

We have,

$\sin^{-1} \left[ \cos \left( \frac{33\pi}{5} \right) \right] = \sin^{-1} \left[ \cos \left( 6\pi + \frac{3\pi}{5} \right) \right] = \sin^{-1} \left[ \cos \left( \frac{3\pi}{5} \right) \right] \quad [∵ \cos(2n\pi + \theta) = \cos \theta]$

$= \sin^{-1} \left[ \cos \left( \frac{\pi}{2} + \frac{\pi}{10} \right) \right] = \sin^{-1} \left( -\sin \frac{\pi}{10} \right) \quad \left[ ∵\cos \left( \frac{\pi}{2} + \theta \right) = -\sin \theta \right]$

$= \sin^{-1} \left( \sin \left( -\frac{\pi}{10} \right) \right) \quad [∵\sin(-x) = -\sin x]$

$= -\frac{\pi}{10} \quad \left[ ∵\sin^{-1}(\sin x) = x, x \in \left[ \frac{-\pi}{2}, \frac{\pi}{2} \right] \right]$