Find the principal value of

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

We know that the range of the principal value branch of $\tan ^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $\tan \left(-\frac{\pi}{3}\right)$ is $-\sqrt{3}$.

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