The value of \(\sec^{-1}\left(\displaystyle\frac{2}{\sqrt{3}}\right) + \tan^{-1}(-1)\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-\displaystyle\frac{\pi}{12}\)
The principal value branch of \(\sec^{-1}\) is \([0, \pi] - \left\{\frac{\pi}{2}\right\}\). Since \(\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\),
\( \displaystyle \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6} \)
The branch of \(\tan^{-1}\) is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). For a negative number the angle is negative:
\( \displaystyle \tan^{-1}(-1) = -\frac{\pi}{4} \)
Adding:
\( \displaystyle \frac{\pi}{6} - \frac{\pi}{4} = \frac{2\pi - 3\pi}{12} \)
\( \displaystyle = -\frac{\pi}{12} \)
Why the other options are incorrect
Option (1) \(\frac{5\pi}{12}\) — Incorrect. This is \(\frac{\pi}{6} + \frac{\pi}{4}\): it takes \(\tan^{-1}(-1)\) as \(+\frac{\pi}{4}\), but \(\tan\frac{\pi}{4} = +1\).
Option (2) \(\frac{\pi}{12}\) — Incorrect. This is \(\frac{\pi}{4} - \frac{\pi}{6}\): the signs of the two principal values are swapped.
Option (3) \(-\frac{5\pi}{12}\) — Incorrect. This is \(-\frac{\pi}{6} - \frac{\pi}{4}\): it takes \(\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\) as \(-\frac{\pi}{6}\), but negative angles are outside \([0, \pi]\).