Question

Find the value of ${\tan }^{-1}\left[2\cos \right(2{\sin }^{-1}\frac{1}{2}\left)\right]$.

Answer 1

Let ${\sin }^{-1}\frac{1}{2}=x$

$\sin x=\frac{1}{2}=\sin \left(\frac{\pi }{6}\right)$

${\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$

Therefore,

${\tan }^{-1}\left[2\cos \right(2{\sin }^{-1}\frac{1}{2}\left)\right]$

$={\tan }^{-1}\left[2\cos \frac{\pi }{3}\right]$

$={\tan }^{-1}1=\frac{\pi }{4}$