Match the LIST-I with LIST-II
| LIST-I (Functions) | LIST-II (Principal Values) | ||
| A. | \( \sin^{-1}\left(-\displaystyle\frac{1}{2}\right)+\cos^{-1}\left(-\displaystyle\frac{1}{2}\right) \) | I. | \( -\displaystyle\frac{\pi}{3} \) |
| B. | \( \cos^{-1}\left(\displaystyle\frac{1}{2}\right)+2\sin^{-1}\left(\displaystyle\frac{1}{2}\right) \) | II. | \( \displaystyle\frac{3\pi}{4} \) |
| C. | \( \tan^{-1}\sqrt{3}-\sec^{-1}(-2) \) | III. | \( \displaystyle\frac{\pi}{2} \) |
| D. | \( \cos^{-1}\left(-\displaystyle\frac{1}{\sqrt{2}}\right) \) | 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
Within the principal value ranges, \( \sin^{-1}(-x)=-\sin^{-1}x \), \( \cos^{-1}(-x)=\pi-\cos^{-1}x \) and \( \sec^{-1}(-x)=\pi-\sec^{-1}x \).
(A) \( \sin^{-1}\left(-\frac{1}{2}\right)+\cos^{-1}\left(-\frac{1}{2}\right) \)
This matches (III) \( \frac{\pi}{2} \).
\( \displaystyle -\frac{\pi}{6}+\left(\pi-\frac{\pi}{3}\right)=-\frac{\pi}{6}+\frac{2\pi}{3} \)
\( \displaystyle =\frac{\pi}{2} \)
(this is also \( \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2} \) with \( x=-\frac{1}{2} \)).
(B) \( \cos^{-1}\left(\frac{1}{2}\right)+2\sin^{-1}\left(\frac{1}{2}\right) \)
This matches (IV) \( \frac{2\pi}{3} \).
\( \displaystyle \frac{\pi}{3}+2\cdot\frac{\pi}{6}=\frac{2\pi}{3} \)
(C) \( \tan^{-1}\sqrt{3}-\sec^{-1}(-2) \)
This matches (I) \( -\frac{\pi}{3} \).
\( \displaystyle \frac{\pi}{3}-\left(\pi-\frac{\pi}{3}\right)=\frac{\pi}{3}-\frac{2\pi}{3} \)
\( \displaystyle =-\frac{\pi}{3} \)
(D) \( \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \)
This matches (II) \( \frac{3\pi}{4} \).
\( \displaystyle \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\pi-\frac{\pi}{4} \)
\( \displaystyle =\frac{3\pi}{4} \)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).
Tip: A negative argument goes to \( \pi- \) for \( \cos^{-1} \) and \( \sec^{-1} \), but to a minus sign for \( \sin^{-1} \) and \( \tan^{-1} \).