The value of the expression $2 \sec^{-1} 2 + \sin^{-1} \left( \frac{1}{2} \right)$ is

$\frac{5\pi}{6}$

The correct answer is Option (2) → $\frac{5\pi}{6}$ ##

We have, $2 \sec^{-1} 2 + \sin^{-1} \frac{1}{2} = 2 \sec^{-1} \left( \sec \frac{\pi}{3} \right) + \sin^{-1} \left( \sin \frac{\pi}{6} \right)$

$\left[ ∵\sec \frac{\pi}{3} = 2 \text{ and } \sin \frac{\pi}{6} = \frac{1}{2} \right]$

$= 2 \cdot \frac{\pi}{3} + \frac{\pi}{6} \quad [∵\sec^{-1}(\sec x) = x \text{ and } \sin^{-1}(\sin x) = x]$

$= \frac{4\pi + \pi}{6} = \frac{5\pi}{6}$