Question

The value of ${\sin }^{-1}\left[\cos \left({\sin }^{-1}\frac{\sqrt{3}}{2}\right)\right]$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{-\pi }{3}$
• D. $\frac{-\pi }{6}$

Answer 1

Answer: (B)

Let $\theta ={\sin }^{-1}\left[\cos \left({\sin }^{-1}\frac{\sqrt{3}}{2}\right)\right]$
$={\sin }^{-1}\left[\cos \frac{\pi }{3}\right]\left(\because \text{ range of }{\sin }^{-1}x\text{ is }\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\right)$
$={\sin }^{-1}\left[\frac{1}{2}\right]=\frac{\pi }{6}$