The value of \(\cos^{-1}\left(\sin\left(-\displaystyle\frac{\pi}{6}\right)\right) + \tan^{-1}(\cos \pi)\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{5\pi}{12}\)
The inner values are
\( \displaystyle \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \)
\(\cos \pi = -1\)
The principal value branch of \(\cos^{-1}\) is \([0, \pi]\). The angle in \([0, \pi]\) whose cosine is \(-\frac{1}{2}\) lies in the second quadrant:
\( \displaystyle \cos^{-1}\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)
The principal value branch of \(\tan^{-1}\) is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). The angle in it whose tangent is \(-1\) is negative:
\( \displaystyle \tan^{-1}(-1) = -\frac{\pi}{4} \)
Adding:
\( \displaystyle \frac{2\pi}{3} - \frac{\pi}{4} = \frac{8\pi - 3\pi}{12} \)
\( \displaystyle = \frac{5\pi}{12} \)
Why the other options are incorrect
Option (1) \(\frac{\pi}{12}\) — Incorrect. This is \(\frac{\pi}{3} - \frac{\pi}{4}\): it takes \(\cos^{-1}\left(-\frac{1}{2}\right)\) as \(\frac{\pi}{3}\), but \(\cos\frac{\pi}{3} = +\frac{1}{2}\).
Option (2) \(\frac{11\pi}{12}\) — Incorrect. This is \(\frac{2\pi}{3} + \frac{\pi}{4}\): it takes \(\tan^{-1}(-1)\) as \(+\frac{\pi}{4}\), but \(\tan\frac{\pi}{4} = +1\).
Option (3) \(\frac{17\pi}{12}\) — Incorrect. This is \(\frac{2\pi}{3} + \frac{3\pi}{4}\): \(\tan\frac{3\pi}{4} = -1\), but \(\frac{3\pi}{4}\) is outside the principal branch \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).