Match List-I with List-II
| List-I | List-II | ||
| (Function) | (Principal value of y) | ||
| (A) | \( y=\sin^{-1}\left(-\displaystyle\frac{1}{2}\right) \) | (I) | \( \displaystyle\frac{3\pi}{4} \) |
| (B) | \( y=\cot^{-1}\left(-\displaystyle\frac{1}{\sqrt{3}}\right) \) | (II) | \( -\displaystyle\frac{\pi}{4} \) |
| (C) | \( y=\mathrm{cosec}^{-1}(-\sqrt{2}) \) | (III) | \( -\displaystyle\frac{\pi}{6} \) |
| (D) | \( y=\cos^{-1}\left(-\displaystyle\frac{1}{\sqrt{2}}\right) \) | (IV) | \( \displaystyle\frac{2\pi}{3} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
Principal value branches: \( \sin^{-1} \) and \( \mathrm{cosec}^{-1} \) in \( \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \); \( \cos^{-1} \) and \( \cot^{-1} \) in \( [0,\pi] \) (open at 0 and π for \( \cot^{-1} \)). So a negative argument gives a negative angle for \( \sin^{-1} \), \( \mathrm{cosec}^{-1} \), but an angle \( \pi-\alpha \) for \( \cos^{-1} \), \( \cot^{-1} \).
(A) \( y=\sin^{-1}\left(-\frac{1}{2}\right) \)
This matches (III) \( -\frac{\pi}{6} \).
\( \displaystyle \sin^{-1}\left(-\frac{1}{2}\right)=-\sin^{-1}\frac{1}{2}=-\frac{\pi}{6} \)
(B) \( y=\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) \)
This matches (IV) \( \frac{2\pi}{3} \).
\( \displaystyle \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)=\pi-\cot^{-1}\frac{1}{\sqrt{3}} \)
\( \displaystyle =\pi-\frac{\pi}{3} \)
\( \displaystyle =\frac{2\pi}{3} \)
(C) \( y=\mathrm{cosec}^{-1}(-\sqrt{2}) \)
This matches (II) \( -\frac{\pi}{4} \).
\( \displaystyle \mathrm{cosec}^{-1}(-\sqrt{2})=-\mathrm{cosec}^{-1}\sqrt{2}=-\frac{\pi}{4} \)
(D) \( y=\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \)
This matches (I) \( \frac{3\pi}{4} \).
\( \displaystyle \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\pi-\cos^{-1}\frac{1}{\sqrt{2}} \)
\( \displaystyle =\pi-\frac{\pi}{4} \)
\( \displaystyle =\frac{3\pi}{4} \)
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).