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

$\frac{\pi}{2}$

The correct answer is Option (2) → $\frac{\pi}{2}$ ##

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

$= \tan^{-1} \left[ 2 \cos \left( 2 \times \frac{\pi}{6} \right) \right] + \frac{\pi}{4}$

$= \tan^{-1} \left( 2 \cos \frac{\pi}{3} \right) + \frac{\pi}{4}$

$= \tan^{-1} \left( 2 \times \frac{1}{2} \right) + \frac{\pi}{4}$

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

$= \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$