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

$\frac{\pi}{6}$ $\frac{\pi}{4}$ $\frac{2\pi}{3}$ $\frac{\pi}{3}$

$\frac{\pi}{4}$

The correct answer is Option (2) → $\frac{\pi}{4}$ $\cot^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2}\right)\right]$ $=\cot^{-1}\left(2\cos\left(2×\frac{π}{6}\right)\right)$ $=\cot^{-1}(2×\frac{1}{2})=\cot^{-1}(1)$ $=\frac{\pi}{4}$