SOLUTION: $\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=\cos ^{-1}\left[\cos \left(\pi-\frac{\pi}{3}\right)\right]+\sin ^{-1}\left[\sin \left(\pi-\frac{\pi}{3}\right)\right]$
$=\cos ^{-1}\left[-\cos \frac{\pi}{3}\right]+\sin ^{-1}\left[\sin \frac{\pi}{3}\right][\because(\pi-x)=-\cos x \text { and } \sin (\pi-x)=\sin x] \\$
$=\pi-\cos ^{-1}\left(\cos \frac{\pi}{3}\right)+\frac{\pi}{3}=\pi-\frac{\pi}{3}+\frac{\pi}{3}=\pi$
