Question

The principal value of ${\sec }^{-1}\frac{2}{\sqrt{3}}$ is ..$A$.. Here, $A$ refers to

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{-\pi }{3}$
• D. $\frac{2\pi }{3}$

Answer 1

Answer: (B)

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