The principal value of \(\tan^{-1}\left(-\displaystyle\frac{1}{\sqrt{3}}\right)\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-\displaystyle\frac{\pi}{6}\)
The principal value branch of \(\tan^{-1}\) is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). For a negative number the angle is negative.
Since \(\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}\) and tangent is an odd function,
\( \displaystyle \tan\left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} \)
and \(-\frac{\pi}{6}\) lies in the branch. Hence \(\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}\).
Why the other options are incorrect
Option (1) \(-\frac{\pi}{3}\) — Incorrect. \(\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}\), not \(-\frac{1}{\sqrt{3}}\).
Option (2) \(\frac{5\pi}{6}\) — Incorrect. \(\tan\frac{5\pi}{6} = -\frac{1}{\sqrt{3}}\), but \(\frac{5\pi}{6}\) is greater than \(\frac{\pi}{2}\), so it is outside the principal branch.
Option (3) \(\frac{\pi}{6}\) — Incorrect. \(\tan\frac{\pi}{6} = +\frac{1}{\sqrt{3}}\); the sign is wrong.