Answer:
Option A
Explanation:
We have,
$\cos ^{-1}\left(\cot\left(\frac{\pi}{2}\right)\right)+\cos ^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$
=$\cos ^{-1}(0)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
$\left[\because \cot \frac{\pi}{2}=0 and \sin \frac{2 \pi}{3}=\frac{\sqrt{3}}{2}\right]$
=$cos ^{-1}\left( \cos \frac{\pi}{2}\right)+cos ^{-1}\left( \cos \frac{\pi}{6}\right)$
$\left[\because \cos \frac{\pi}{2}=0 and \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right]$
= $\frac{\pi}{2}+\frac{\pi}{6}$
$=\frac{3\pi+\pi}{6}=\frac{4\pi}{6}=\frac{2 \pi}{3}$
