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