Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(\sin^{-1}(-1)\) | (I) | \(\displaystyle\frac{5\pi}{6}\) |
| (B) | \(\cot^{-1}(-1)\) | (II) | \(\displaystyle\frac{-\pi}{2}\) |
| (C) | \(\sec^{-1}\left(\displaystyle\frac{-2}{\sqrt{3}}\right)\) | (III) | \(\displaystyle\frac{\pi}{4}\) |
| (D) | \(\tan^{-1}(1)\) | (IV) | \(\displaystyle\frac{3\pi}{4}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Each value must lie in the principal value branch of its function:
\(\sin^{-1}\): \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
\(\cot^{-1}\): \((0, \pi)\)
\(\sec^{-1}\): \([0, \pi] - \left\{\frac{\pi}{2}\right\}\)
\(\tan^{-1}\): \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
(A) \(\sin^{-1}(-1)\)
This matches (II) \(\frac{-\pi}{2}\), since \(\sin\left(-\frac{\pi}{2}\right) = -1\) and \(-\frac{\pi}{2}\) is in the branch.
(B) \(\cot^{-1}(-1)\)
This matches (IV) \(\frac{3\pi}{4}\):
\( \displaystyle \cot^{-1}(-1) = \pi - \cot^{-1}(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)
(C) \(\sec^{-1}\left(\frac{-2}{\sqrt{3}}\right)\)
This matches (I) \(\frac{5\pi}{6}\). Since \(\sec^{-1} x = \cos^{-1}\frac{1}{x}\):
\( \displaystyle \sec^{-1}\left(\frac{-2}{\sqrt{3}}\right) = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \)
\( \displaystyle = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \)
(D) \(\tan^{-1}(1)\)
This matches (III) \(\frac{\pi}{4}\), since \(\tan\frac{\pi}{4} = 1\) and \(\frac{\pi}{4}\) is in the branch.
So the matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).
Tip: For a negative argument, \(\cot^{-1}(-x) = \pi - \cot^{-1}x\) and \(\sec^{-1}(-x) = \pi - \sec^{-1}x\) (as for \(\cos^{-1}\)). Only \(\sin^{-1}\), \(\tan^{-1}\) and \(\mathrm{cosec}^{-1}\) give negative angles.