Solution

We know that the principal range of $\cos^{-1}x$ is $[0,\pi]$.

Now,

$\cos\left(-\frac{\pi}{6}\right)=\cos\frac{\pi}{6}$

So,

$\cos^{-1}\left(\cos\left(-\frac{\pi}{6}\right)\right)=\frac{\pi}{6}$

Also,

$\sin\frac{5\pi}{6}=\frac{1}{2}$

The principal range of $\sin^{-1}x$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

So,

$\sin^{-1}\left(\sin\frac{5\pi}{6}\right)=\sin^{-1}\left(\frac{1}{2}\right)$

$=\frac{\pi}{6}$

Therefore,

$\cos^{-1}\left(\cos\left(-\frac{\pi}{6}\right)\right)+\sin^{-1}\left(\sin\frac{5\pi}{6}\right)$

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

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