Find the principal value of sec ^{-1}left(fra | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

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

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

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