Step 1: Remember the safe ranges.
$\cos^{-1}$ always returns an angle in $[0,\pi]$, and $\sin^{-1}$ always returns an angle in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. The trick is to replace the inside angle with an equivalent angle that lives in these ranges.
Step 2: Fix the first term using evenness of cosine.
Since $\cos$ is even, $\cos\left(-\frac{\pi}{6}\right)=\cos\left(\frac{\pi}{6}\right)$. The angle $\frac{\pi}{6}$ already sits inside $[0,\pi]$.
Step 3: Collapse the first term.
Because $\frac{\pi}{6}$ is in the valid range, $\cos^{-1}\left(\cos\frac{\pi}{6}\right)=\frac{\pi}{6}$ directly.
Step 4: Find a friendly angle for the second term.
$\sin\left(\frac{5\pi}{6}\right)=\sin\left(\pi-\frac{5\pi}{6}\right)=\sin\left(\frac{\pi}{6}\right)$, and $\frac{\pi}{6}$ lies in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.
Step 5: Collapse the second term.
So $\sin^{-1}\left(\sin\frac{5\pi}{6}\right)=\sin^{-1}\left(\sin\frac{\pi}{6}\right)=\frac{\pi}{6}$.
Step 6: Add the two clean values.
$\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}$, which is option 2.
\[ \boxed{\frac{\pi}{3}} \]