Question

Find the principal value of ${\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)$.

Answer 1

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

$\Rightarrow \sec y=\frac{2}{\sqrt{3}}=\sec \left(\frac{\pi }{6}\right)$.

We know that the range of the principal value of ${\sec }^{-1}x$ 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 $\frac{\pi }{6}.$