The value of \(\tan^{-1}(1)+\cos^{-1}\left(-\displaystyle\frac{\sqrt{3}}{2}\right) + \sin^{-1}\left(\displaystyle\frac{1}{2}\right)\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{5\pi}{4}\)
Each value is taken in its principal value branch: \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) for \(\tan^{-1}\), \([0, \pi]\) for \(\cos^{-1}\) and \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) for \(\sin^{-1}\).
\(\displaystyle \tan^{-1}(1) = \frac{\pi}{4}\)
\(\displaystyle \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \pi - \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\)
\(\displaystyle \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\)
Adding:
\(\displaystyle \frac{\pi}{4} + \frac{5\pi}{6} + \frac{\pi}{6} = \frac{\pi}{4} + \pi = \frac{5\pi}{4}\)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{\pi}{4}\) — Incorrect. This comes from taking \(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{6}\), which lies outside the principal range \([0, \pi]\) of \(\cos^{-1}\).
Option (3) \(\displaystyle\frac{5\pi}{12}\) — Incorrect. The three values add up to \(\frac{5\pi}{4}\), not \(\frac{5\pi}{12}\).
Option (4) \(\displaystyle\frac{7\pi}{12}\) — Incorrect. This comes from taking \(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}\), ignoring the minus sign: \(\frac{\pi}{4} + \frac{\pi}{6} + \frac{\pi}{6} = \frac{7\pi}{12}\).
Tip: For a negative argument, \(\cos^{-1}(-x) = \pi - \cos^{-1}x\), whereas \(\sin^{-1}(-x) = -\sin^{-1}x\) and \(\tan^{-1}(-x) = -\tan^{-1}x\). Mixing up these two rules is a common mistake.