Solution:

Option(D) is correct.
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 $\cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)\left(\because\right.$ the principle value of $\cos$ lies in the range $[0, \pi]$ and since $\left.\frac{2 \pi}{3} \in[0, \pi]\right)$ $\Rightarrow \cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)=\frac{2 \pi}{3}$
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 $\frac{2 \pi}{3} \notin\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\begin{array}{l}
\Rightarrow \sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)=\sin ^{-1}\left(\sin \left(\pi-\frac{\pi}{3}\right)\right) \\
=\sin ^{-1}\left(\sin \left(\frac{\pi}{3}\right)\right) \\
=\frac{\pi}{3}
\end{array}$
Therefore,
$\cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)+\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)=\frac{2 \pi}{3}+\frac{\pi}{3}$
$=\frac{3 \pi}{3}$
$=\pi$