Question

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

Answer 1

Let $y={\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)$
$\Rightarrow \sec y=\frac{2}{\sqrt{3}}$
$\Rightarrow \sec y=\sec \left(\frac{\pi }{6}\right)$
We know that the range of the principal value branch of ${\sec }^{-1}x$ is $[0,\pi ]-\left\{\frac{\pi }{2}\right\}$.
$\therefore y=\frac{\pi }{6}$
Hence, the principal value of ${\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is $\frac{\pi }{6}.$