The value of \(2\sin^{-1}\left(\displaystyle\frac{\sqrt{3}}{2}\right)+\sec^{-1}\left(\displaystyle\frac{2}{\sqrt{3}}\right)+\cot^{-1}\left(\displaystyle\frac{-1}{\sqrt{3}}\right)\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{3\pi}{2}\)
Each term takes its principal value.
The principal range of \(\sin^{-1}\) is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\):
\( \displaystyle 2\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 2\times\frac{\pi}{3} = \frac{2\pi}{3} \)
The principal range of \(\sec^{-1}\) is \([0, \pi]\) without \(\frac{\pi}{2}\), and \(\sec\frac{\pi}{6} = \frac{2}{\sqrt{3}}\):
\( \displaystyle \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6} \)
The principal range of \(\cot^{-1}\) is \((0, \pi)\), so \(\cot^{-1}(-x) = \pi - \cot^{-1}x\):
\( \displaystyle \cot^{-1}\left(\frac{-1}{\sqrt{3}}\right) = \pi - \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) \)
\( \displaystyle = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)
Adding:
\( \displaystyle \frac{2\pi}{3}+\frac{\pi}{6}+\frac{2\pi}{3} = \frac{4\pi+\pi+4\pi}{6} \)
\( \displaystyle = \frac{9\pi}{6} = \frac{3\pi}{2} \)
Tip: For a negative argument, \(\cot^{-1}(-x) = \pi - \cot^{-1}x\), not \(-\cot^{-1}x\), because the principal value of \(\cot^{-1}\) always lies in (0, π). The same holds for \(\cos^{-1}\) and \(\sec^{-1}\).