Match List-I with List-II
|
List-I (Inverse Trigonometric Function) |
List-II (Principal Value) |
|
(A) $\sin^{-1}(-\frac{1}{2})$ |
(I) $π/6$ |
|
(B) $\cos^{-1}(-\frac{1}{2})$ |
(II) $-π/6$ |
|
(C) $\tan^{-1}(-\sqrt{3})$ |
(III) $2π/3$ |
|
(D) $\cot^{-1}(\sqrt{3})$ |
(IV) $-π/3$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
|
List-I (Inverse Trigonometric Function) |
List-II (Principal Value) |
|
(A) $\sin^{-1}(-\frac{1}{2})$ |
(II) $-π/6$ |
|
(B) $\cos^{-1}(-\frac{1}{2})$ |
(III) $2π/3$ |
|
(C) $\tan^{-1}(-\sqrt{3})$ |
(IV) $-π/3$ |
|
(D) $\cot^{-1}(\sqrt{3})$ |
(I) $π/6$ |
(A) $\sin^{-1}\left(-\frac{1}{2}\right)$ Principal value lies in $[-\frac{\pi}{2},\frac{\pi}{2}]$. $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$ → (A) → (II)
(B) $\cos^{-1}\left(-\frac{1}{2}\right)$ Principal value lies in $[0,\pi]$. $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ → (B) → (III)
(C) $\tan^{-1}(-\sqrt{3})$ Principal value lies in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. $\tan^{-1}(\sqrt{3})=\frac{\pi}{3}$ → negative gives $\tan^{-1}(-\sqrt{3})=-\frac{\pi}{3}$ → (C) → (IV)
(D) $\cot^{-1}(\sqrt{3})$ Principal value lies in $(0,\pi)$. $\cot^{-1}(\sqrt{3})=\frac{\pi}{6}$ → (D) → (I)
Final matching:
(A)–(II), (B)–(III), (C)–(IV), (D)–(I)