Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(\cos^{-1}x + \cos^{-1}(-x)\) | (I) | \(\displaystyle\frac{\pi}{3}\) |
| (B) | \(\mathrm{cosec}^{-1}(-x) + \sec^{-1}(-x)\) | (II) | \(-\displaystyle\frac{\pi}{3}\) |
| (C) | \(\tan^{-1}\sqrt{3} - \sec^{-1}(-2)\) | (III) | \(\pi\) |
| (D) | \(\tan^{-1}\left(\tan\displaystyle\frac{4\pi}{3}\right)\) | (IV) | \(\displaystyle\frac{\pi}{2}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(A) \(\cos^{-1}x + \cos^{-1}(-x)\)
This matches (III) \(\pi\). Using \(\cos^{-1}(-x) = \pi - \cos^{-1}x\):
\(\cos^{-1}x + \pi - \cos^{-1}x = \pi\)
(B) \(\mathrm{cosec}^{-1}(-x) + \sec^{-1}(-x)\)
This matches (IV) \(\frac{\pi}{2}\). For |x| ≥ 1, \(\mathrm{cosec}^{-1}(-x) = -\mathrm{cosec}^{-1}x\) and \(\sec^{-1}(-x) = \pi - \sec^{-1}x\):
\(\mathrm{cosec}^{-1}(-x) + \sec^{-1}(-x) = \pi - (\mathrm{cosec}^{-1}x + \sec^{-1}x)\)
\(\displaystyle = \pi - \frac{\pi}{2} = \frac{\pi}{2}\)
(C) \(\tan^{-1}\sqrt{3} - \sec^{-1}(-2)\)
This matches (II) \(-\frac{\pi}{3}\):
\(\displaystyle \tan^{-1}\sqrt{3} = \frac{\pi}{3}\)
\(\displaystyle \sec^{-1}(-2) = \pi - \sec^{-1}2 = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\)
\(\displaystyle \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}\)
(D) \(\tan^{-1}\left(\tan\frac{4\pi}{3}\right)\)
This matches (I) \(\frac{\pi}{3}\). The principal value of \(\tan^{-1}\) lies in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), and \(\frac{4\pi}{3}\) is outside it. Since tan has period π:
\(\displaystyle \tan\frac{4\pi}{3} = \tan\left(\pi + \frac{\pi}{3}\right) = \tan\frac{\pi}{3}\)
\(\displaystyle \tan^{-1}\left(\tan\frac{\pi}{3}\right) = \frac{\pi}{3}\)
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).