Question: Find the principal value of \begin{align} sec^{-1}\left(\frac{2}{\sqrt3}\right)\end{align}

Answer:

\begin{align} Let \;\; sec^{-1}\left(\frac{2}{\sqrt3}\right)=y \;\;Then\;\; sec y = \frac{2}{\sqrt3} = sec\left(\frac{\pi}{6}\right)\end{align}

We know that the range of the principal value branch of sec⁻¹ is

\begin{align} \left[0,\pi\right] - \left(\frac{\pi}{2}\right) and \;\;sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt3}\end{align}

Therefore, the principal value of

\begin{align} sec^{-1}\left(\frac{2}{\sqrt3}\right) is \frac{\pi}{6}\end{align}

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