The value of \(\sin^{-1}\left(\displaystyle\frac{\sqrt{3}}{2}\right) - \cot^{-1}\left(-\sqrt{3}\right)\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-\displaystyle\frac{\pi}{2}\)
The principal value branch of \(\sin^{-1}\) is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). Since \(\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\),
\( \displaystyle \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \)
The principal value branch of \(\cot^{-1}\) is \((0, \pi)\). A negative cotangent needs an angle between \(\frac{\pi}{2}\) and \(\pi\). Since \(\cot\frac{\pi}{6} = \sqrt{3}\),
\( \displaystyle \cot\left(\pi - \frac{\pi}{6}\right) = -\sqrt{3} \)
\( \displaystyle \cot^{-1}\left(-\sqrt{3}\right) = \frac{5\pi}{6} \)
Subtracting:
\( \displaystyle \frac{\pi}{3} - \frac{5\pi}{6} = \frac{2\pi - 5\pi}{6} \)
\( \displaystyle = -\frac{\pi}{2} \)
Why the other options are incorrect
Option (1) \(\frac{\pi}{2}\) — Incorrect. This is \(\frac{\pi}{3} + \frac{\pi}{6}\): it takes \(\cot^{-1}\left(-\sqrt{3}\right)\) as \(-\frac{\pi}{6}\), but negative angles are outside \((0, \pi)\).
Option (2) \(\frac{\pi}{6}\) — Incorrect. This is \(\frac{\pi}{3} - \frac{\pi}{6}\): it takes \(\cot^{-1}\left(-\sqrt{3}\right)\) as \(\frac{\pi}{6}\), but \(\cot\frac{\pi}{6} = +\sqrt{3}\).
Option (3) \(-\frac{2\pi}{3}\) — Incorrect. This is \(\frac{\pi}{6} - \frac{5\pi}{6}\): it takes \(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\) as \(\frac{\pi}{6}\), but \(\sin\frac{\pi}{6} = \frac{1}{2}\).