Match List – I with List – II
| List – I | List – II | ||
| A. | \(\tan^{-1}\left(\tan\displaystyle\frac{7\pi}{6}\right)\) | I. | \(\displaystyle\frac{5\pi}{6}\) |
| B. | \(\tan^{-1}\left(\tan\displaystyle\frac{8\pi}{6}\right)\) | II. | \(\displaystyle\frac{\pi}{2}\) |
| C. | \(\tan^{-1}\displaystyle\frac{1}{\sqrt{3}}+\mathrm{cosec}^{-1}\frac{2}{\sqrt{3}}\) | III. | \(\displaystyle\frac{\pi}{6}\) |
| D. | \(\cos^{-1}\left(\cos\displaystyle\frac{5\pi}{6}\right)\) | IV. | \(\displaystyle\frac{\pi}{3}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-IV, C-II, D-I
The principal value branches are \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) for \(\tan^{-1}\), \([0, \pi]\) for \(\cos^{-1}\) and \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) without 0 for \(\mathrm{cosec}^{-1}\). The answer must lie in the right branch.
(A) \(\tan^{-1}\left(\tan\frac{7\pi}{6}\right)\)
This matches (III) \(\frac{\pi}{6}\). Since tan has period \(\pi\):
\( \displaystyle \tan\frac{7\pi}{6} = \tan\left(\pi + \frac{\pi}{6}\right) = \tan\frac{\pi}{6} \)
and \(\frac{\pi}{6}\) lies in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
(B) \(\tan^{-1}\left(\tan\frac{8\pi}{6}\right)\)
This matches (IV) \(\frac{\pi}{3}\).
\( \displaystyle \tan\frac{8\pi}{6} = \tan\left(\pi + \frac{\pi}{3}\right) = \tan\frac{\pi}{3} \)
and \(\frac{\pi}{3}\) lies in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
(C) \(\tan^{-1}\frac{1}{\sqrt{3}}+\mathrm{cosec}^{-1}\frac{2}{\sqrt{3}}\)
This matches (II) \(\frac{\pi}{2}\). Since \(\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}\) and \(\mathrm{cosec}\,\frac{\pi}{3} = \frac{2}{\sqrt{3}}\):
\( \displaystyle \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2} \)
(D) \(\cos^{-1}\left(\cos\frac{5\pi}{6}\right)\)
This matches (I) \(\frac{5\pi}{6}\), because \(\frac{5\pi}{6}\) already lies in \([0, \pi]\).
So the matching is A-III, B-IV, C-II, D-I.