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To Find: The value of $\cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)+\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$

Here,consider $\cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)$ (\because the principle value of $\cos$ lies in the range $[0, \pi]$ and since $\frac{2 \pi}{3} \in[0, \pi]$ )

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

Now, consider $\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$

Since here the principle value of sine lies in range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and since $\frac{2 \pi}{3} \notin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

$\Rightarrow \sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)=\sin ^{-1}\left(\sin \left(\pi-\frac{\pi}{3}\right)\right)$

$=\sin ^{-1}\left(\sin \left(\frac{\pi}{3}\right)\right)$

$=\frac{\pi}{3}$

Therefore,

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

$=\frac{3 \pi}{3}$

$=\pi$