Consider the following statements:
(A) \(\sin^{-1}\left(\sin\displaystyle\frac{2\pi}{3}\right) = \displaystyle\frac{2\pi}{3}\)
(B) \(\cos^{-1}\left(\cos\left(-\displaystyle\frac{\pi}{4}\right)\right) = \displaystyle\frac{\pi}{4}\)
(C) \(\tan^{-1}\left(\tan\left(-\displaystyle\frac{\pi}{3}\right)\right) = -\displaystyle\frac{\pi}{3}\)
(D) \(\cot^{-1}\left(-\sqrt{3}\right) = -\displaystyle\frac{\pi}{6}\)
Which of the above statements are correct?
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B) and (C) only
An inverse function returns only values in its principal value branch: \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) for \(\sin^{-1}\), \([0, \pi]\) for \(\cos^{-1}\), \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) for \(\tan^{-1}\), \((0, \pi)\) for \(\cot^{-1}\).
\( \displaystyle \sin\frac{2\pi}{3} = \sin\left(\pi - \frac{2\pi}{3}\right) = \sin\frac{\pi}{3} \)
\( \displaystyle \cos\left(-\frac{\pi}{4}\right) = \cos\frac{\pi}{4} \)
\( \displaystyle \cot\left(\pi - \frac{\pi}{6}\right) = -\cot\frac{\pi}{6} = -\sqrt{3} \)
So \(\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \frac{\pi}{3}\), \(\cos^{-1}\left(\cos\left(-\frac{\pi}{4}\right)\right) = \frac{\pi}{4}\), \(\tan^{-1}\left(\tan\left(-\frac{\pi}{3}\right)\right) = -\frac{\pi}{3}\) and \(\cot^{-1}\left(-\sqrt{3}\right) = \frac{5\pi}{6}\).
Checking the statements
(A) \(\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \frac{2\pi}{3}\) — Incorrect. The value is \(\frac{\pi}{3}\); \(\frac{2\pi}{3}\) is outside \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
(B) \(\cos^{-1}\left(\cos\left(-\frac{\pi}{4}\right)\right) = \frac{\pi}{4}\) — Correct. \(\cos\left(-\frac{\pi}{4}\right) = \cos\frac{\pi}{4}\) and \(\frac{\pi}{4}\) is in \([0, \pi]\).
(C) \(\tan^{-1}\left(\tan\left(-\frac{\pi}{3}\right)\right) = -\frac{\pi}{3}\) — Correct. \(-\frac{\pi}{3}\) is already in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
(D) \(\cot^{-1}\left(-\sqrt{3}\right) = -\frac{\pi}{6}\) — Incorrect. The value is \(\frac{5\pi}{6}\); negative angles are outside \((0, \pi)\).