The value of \(\cot^{-1}\left[2\sin\left(2\cos^{-1}\displaystyle\frac{\sqrt{3}}{2}\right)\right] + \cos^{-1}\left(-\displaystyle\frac{1}{2}\right)\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{5\pi}{6}\)
We work from the inside out, taking every inverse value in its principal value branch.
\( \displaystyle \cos^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{6} \)
\( \displaystyle 2\sin\left(2 \cdot \frac{\pi}{6}\right) = 2\sin\frac{\pi}{3} \)
\( \displaystyle = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \)
Since \(\cot\frac{\pi}{6} = \sqrt{3}\) and \(\frac{\pi}{6}\) is in \((0, \pi)\),
\( \displaystyle \cot^{-1}\left(\sqrt{3}\right) = \frac{\pi}{6} \)
For a negative number \(\cos^{-1}\) gives a second-quadrant angle:
\( \displaystyle \cos^{-1}\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)
Adding:
\( \displaystyle \frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi + 4\pi}{6} \)
\( \displaystyle = \frac{5\pi}{6} \)
Why the other options are incorrect
Option (1) \(\frac{\pi}{2}\) — Incorrect. This is \(\frac{\pi}{6} + \frac{\pi}{3}\): it takes \(\cos^{-1}\left(-\frac{1}{2}\right)\) as \(\frac{\pi}{3}\), but \(\cos\frac{\pi}{3} = +\frac{1}{2}\).
Option (2) \(-\frac{\pi}{6}\) — Incorrect. This is \(\frac{\pi}{6} - \frac{\pi}{3}\): it takes \(\cos^{-1}\left(-\frac{1}{2}\right)\) as \(-\frac{\pi}{3}\), but \(\cos^{-1}\) never gives a negative angle.
Option (4) \(\pi\) — Incorrect. This is \(\frac{\pi}{3} + \frac{2\pi}{3}\): it takes \(\cot^{-1}\left(\sqrt{3}\right)\) as \(\frac{\pi}{3}\), but \(\cot\frac{\pi}{3} = \frac{1}{\sqrt{3}}\).