Find the principal value of

Let $\cot ^{-1}(\sqrt{3})=y$. Then, $\cot y=\sqrt{3}=\cot \left(\frac{\pi}{6}\right)$.

We know that the range of the principal value branch of $\cot ^{-1}$ is $(0, \pi)$ and

$\cot \left(\frac{\pi}{6}\right)=\sqrt{3}$

Therefore, the principal value of $\cot ^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$.