Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Inverse Trigonometric function | Principal values of arguments | ||
| (A) | \(\sin^{-1}\left(\displaystyle\frac{-1}{2}\right)\) | (I) | \(\displaystyle\frac{-\pi}{3}\) |
| (B) | \(\cos^{-1}\left(\displaystyle\frac{-1}{2}\right)\) | (II) | \(\displaystyle\frac{3\pi}{4}\) |
| (C) | \(\tan^{-1}(-\sqrt{3})\) | (III) | \(\displaystyle\frac{-\pi}{6}\) |
| (D) | \(\sec^{-1}(-\sqrt{2})\) | (IV) | \(\displaystyle\frac{2\pi}{3}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Each value must lie in the principal value branch: \(\sin^{-1}\) in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), \(\cos^{-1}\) in \([0, \pi]\), \(\tan^{-1}\) in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), \(\sec^{-1}\) in \([0, \pi] - \left\{\frac{\pi}{2}\right\}\).
(A) \(\sin^{-1}\left(\frac{-1}{2}\right)\)
This matches (III) \(\frac{-\pi}{6}\), since \(\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\) and \(-\frac{\pi}{6}\) is in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
(B) \(\cos^{-1}\left(\frac{-1}{2}\right)\)
This matches (IV) \(\frac{2\pi}{3}\), since \(\cos^{-1}(-x) = \pi - \cos^{-1}x\):
\(\displaystyle \cos^{-1}\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\)
(C) \(\tan^{-1}(-\sqrt{3})\)
This matches (I) \(\frac{-\pi}{3}\), since \(\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}\) and \(-\frac{\pi}{3}\) is in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
(D) \(\sec^{-1}(-\sqrt{2})\)
This matches (II) \(\frac{3\pi}{4}\), since \(\sec^{-1}(-x) = \pi - \sec^{-1}x\):
\(\displaystyle \sec^{-1}(-\sqrt{2}) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).