The principal value of \(\sec^{-1}\left(-\displaystyle\frac{2}{\sqrt{3}}\right)\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{5\pi}{6}\)
The principal value branch of \(\sec^{-1}\) is \([0, \pi] - \left\{\frac{\pi}{2}\right\}\). A negative secant means a negative cosine, so the angle lies between \(\frac{\pi}{2}\) and \(\pi\).
Since \(\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\),
\( \displaystyle \sec\frac{\pi}{6} = \frac{2}{\sqrt{3}} \)
Moving to the second quadrant changes the sign:
\( \displaystyle \sec\left(\pi - \frac{\pi}{6}\right) = -\frac{2}{\sqrt{3}} \)
Hence \(\sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) = \frac{5\pi}{6}\).
Why the other options are incorrect
Option (1) \(\frac{\pi}{6}\) — Incorrect. \(\sec\frac{\pi}{6} = +\frac{2}{\sqrt{3}}\), not \(-\frac{2}{\sqrt{3}}\).
Option (2) \(-\frac{\pi}{6}\) — Incorrect. Negative angles are outside \([0, \pi]\), and \(\sec\left(-\frac{\pi}{6}\right) = +\frac{2}{\sqrt{3}}\).
Option (4) \(\frac{7\pi}{6}\) — Incorrect. \(\sec\frac{7\pi}{6} = -\frac{2}{\sqrt{3}}\), but \(\frac{7\pi}{6}\) is greater than \(\pi\), so it is outside the principal branch.