Find the principal value of

Let $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=y$. Then, $\sec y=\frac{2}{\sqrt{3}}=\sec \left(\frac{\pi}{6}\right)$.

We know that the range of the principal value branch of $\mathrm{sec}^{-1}$ is

$[0, \pi]-\left\{\frac{\pi}{2}\right\}$ and $\sec \left(\frac{\pi}{6}\right)=\frac{2}{\sqrt{3}}$

Therefore, the principal value of $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is $\frac{\pi}{6}$.