Match List-I with List-II

List-I	List-II
(A) $\sin^{-1}(-1)$	(I) $\frac{5\pi}{6}$
(B) $\cot^{-1}(-1)$	(II) $\frac{-\pi}{2}$
(C) $\sec^{-1}\left(\frac{-2}{\sqrt{3}}\right)$	(III) $\frac{\pi}{4}$
(D) $\tan^{-1}(1)$	(IV) $\frac{3\pi}{4}$

Choose the correct answer from the options given below:

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

Principal range for $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

At angle $\frac{-\pi}{2}$, $\sin\left(\frac{-\pi}{2}\right) = -1$

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

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

Principal range for $\cot^{-1}$ is $(0, \pi)$

Since $\cot(\theta) = -1$, then $\tan(\theta) = -1$

In the range $(0, \pi)$, tangent is negative in the second quadrant.

At angle $\frac{3\pi}{4}$, $\tan\left(\frac{3\pi}{4}\right) = -1$

Therefore $\cot^{-1}(-1) = \frac{3\pi}{4}$ → (IV)

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For $\sec^{-1}\left(\frac{-2}{\sqrt{3}}\right)$:

Principal range for $\sec^{-1}$ is $[0, \pi]$ excluding $\frac{\pi}{2}$

$\sec(\theta) = \frac{-2}{\sqrt{3}}$ means $\cos(\theta) = \frac{-\sqrt{3}}{2}$

Cosine is negative in the second quadrant.

At angle $\frac{5\pi}{6}$, $\cos\left(\frac{5\pi}{6}\right) = \frac{-\sqrt{3}}{2}$

Therefore $\sec^{-1}\left(\frac{-2}{\sqrt{3}}\right) = \frac{5\pi}{6}$ → (I)

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

Principal range for $\tan^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

At angle $\frac{\pi}{4}$, $\tan\left(\frac{\pi}{4}\right) = 1$

Therefore $\tan^{-1}(1) = \frac{\pi}{4}$ → (III)

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Final matching:

Function	Value	Matches
(A) $\sin^{-1}(-1)$	$\frac{-\pi}{2}$	(II)
(B) $\cot^{-1}(-1)$	$\frac{3\pi}{4}$	(IV)
(C) $\sec^{-1}\left(\frac{-2}{\sqrt{3}}\right)$	$\frac{5\pi}{6}$	(I)
(D) $\tan^{-1}(1)$	$\frac{\pi}{4}$	(III)

The correct answer is (A)-(II), (B)-(IV), (C)-(I), (D)-(III)