The value of $tan^{-1}\begin{Bmatrix}2cos\left(2sin^{-1}\frac{1}{2}\right)\end{Bmatrix},$ is
Answer & explanation
Correct answer: option 1
We know that $sin^{-1}\frac{1}{2}=\frac{\pi}{6}.$
$∴ tan^{-1}\begin{Bmatrix}2 cos \left(2sin^{-1}\frac{1}{2}\right)\end{Bmatrix}$
$= tan^{-1}\begin{Bmatrix}2cos \left(2× \frac{\pi}{6}\right)\end{Bmatrix}= tan^{-1}\left(2cos\frac{\pi}{3}\right)$
$=tan^{-1}\left(2 × \frac{1}{2}\right) = tan^{-1} =\frac{\pi}{4}$