Match List-I with List-II

List-I	List-II
(Inverse Trigonometric Function)	(Principal Value)
(A) $\sin^{-1}(-\frac{1}{2})$	(I) ${\pi}/{6}$
(B) $\cos^{-1}(-\frac{1}{2})$	(II) $-{\pi}/{6}$
(C) $\tan^{-1}(-\sqrt{3})$	(III) ${2\pi}/{3}$
(D) $\cot^{-1}(\sqrt{3})$	(IV) $-{\pi}/{3}$

Choose the correct answer from the options given below:

Principal value ranges for inverse trigonometric functions:

$\sin^{-1}(x)$: $[-\frac{\pi}{2}, \frac{\pi}{2}]$

$\cos^{-1}(x)$: $[0, \pi]$

$\tan^{-1}(x)$: $(-\frac{\pi}{2}, \frac{\pi}{2})$

$\cot^{-1}(x)$: $(0, \pi)$

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For $\sin^{-1}(-\frac{1}{2})$:

Need an angle in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ whose sine equals $-\frac{1}{2}$.

Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$ and sine is negative for negative angles:

$\sin(-\frac{\pi}{6}) = -\frac{1}{2}$

Therefore, $\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$ → (II)

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For $\cos^{-1}(-\frac{1}{2})$:

Need an angle in $[0, \pi]$ whose cosine equals $-\frac{1}{2}$.

Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and cosine is negative in the second quadrant:

$\pi - \frac{\pi}{3}$

$= \frac{3\pi - \pi}{3}$

$= \frac{2\pi}{3}$

Therefore, $\cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}$ → (III)

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For $\tan^{-1}(-\sqrt{3})$:

Need an angle in $(-\frac{\pi}{2}, \frac{\pi}{2})$ whose tangent equals $-\sqrt{3}$.

Since $\tan(\frac{\pi}{3}) = \sqrt{3}$:

$\tan(-\frac{\pi}{3}) = -\sqrt{3}$

Therefore, $\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$ → (IV)

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For $\cot^{-1}(\sqrt{3})$:

Need an angle in $(0, \pi)$ whose cotangent equals $\sqrt{3}$.

$\cot(\frac{\pi}{6})$

$= \frac{\cos(\frac{\pi}{6})}{\sin(\frac{\pi}{6})}$

$= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$

$= \sqrt{3}$

Therefore, $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$ → (I)

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Final matching:

Function	Principal Value
(A) $\sin^{-1}(-\frac{1}{2})$	(II) $-\frac{\pi}{6}$
(B) $\cos^{-1}(-\frac{1}{2})$	(III) $\frac{2\pi}{3}$
(C) $\tan^{-1}(-\sqrt{3})$	(IV) $-\frac{\pi}{3}$
(D) $\cot^{-1}(\sqrt{3})$	(I) $\frac{\pi}{6}$

Correct answer: (A)-(II), (B)-(III), (C)-(IV), (D)-(I)