Match List-I with List-II
| List-I | List-II | ||
| Inverse trigonometric function | Principal Value | ||
| (A) | \( \sin^{-1}(-1/2) \) | (I) | \( 5\pi/6 \) |
| (B) | \( \cos^{-1}\left(\displaystyle\frac{-\sqrt{3}}{2}\right) \) | (II) | \( 2\pi/3 \) |
| (C) | \( \tan^{-1}(\sqrt{3}) \) | (III) | \( -\pi/6 \) |
| (D) | \( \sec^{-1}(-2) \) | (IV) | \( \pi/3 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
The principal value branches are \( \sin^{-1} \) in \( \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \), \( \cos^{-1} \) in \( [0,\pi] \), \( \tan^{-1} \) in \( \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \) and \( \sec^{-1} \) in \( [0,\pi]-\left\{\frac{\pi}{2}\right\} \).
(A) \( \sin^{-1}(-1/2) \)
This matches (III) −π/6.
\( \displaystyle \sin^{-1}(-x)=-\sin^{-1}x \)
\( \displaystyle \sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6} \)
(B) \( \cos^{-1}\left(\frac{-\sqrt{3}}{2}\right) \)
This matches (I) 5π/6.
\( \displaystyle \cos^{-1}(-x)=\pi-\cos^{-1}x \)
\( \displaystyle \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\pi-\frac{\pi}{6}=\frac{5\pi}{6} \)
(C) \( \tan^{-1}(\sqrt{3}) \)
This matches (IV) π/3.
\( \displaystyle \tan\frac{\pi}{3}=\sqrt{3}\ \text{and}\ \frac{\pi}{3}\ \text{lies in}\ \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \)
(D) \( \sec^{-1}(-2) \)
This matches (II) 2π/3.
\( \displaystyle \sec^{-1}(-2)=\cos^{-1}\left(-\frac{1}{2}\right) \)
\( \displaystyle =\pi-\frac{\pi}{3}=\frac{2\pi}{3} \)
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).
Tip: For a negative argument, \( \sin^{-1} \), \( \tan^{-1} \) and \( \mathrm{cosec}^{-1} \) just take a minus sign, but \( \cos^{-1} \), \( \sec^{-1} \) and \( \cot^{-1} \) give \( \pi \) minus the value, because their branches lie in \( [0,\pi] \).