Match List-I with List-II

List-I	List-II
(A) $\cos^{-1} x + \cos^{-1}(-x)$	(I) $\frac{\pi}{3}$
(B) $\text{cosec}^{-1}(-x) + \sec^{-1}(-x)$	(II) $-\frac{\pi}{3}$
(C) $\tan^{-1}\sqrt{3} - \sec^{-1}(-2)$	(III) $\pi$
(D) $\tan^{-1}\left(\tan\frac{4\pi}{3}\right)$	(IV) $\frac{\pi}{2}$

Choose the correct answer from the options given below:

For (A), using the property $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$:

$\cos^{-1} x + \cos^{-1}(-x)$

$= \cos^{-1} x + \pi - \cos^{-1} x$

$= \pi$

(A) matches with (III)

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For (B), using $\text{cosec}^{-1}(-x) = -\text{cosec}^{-1}(x)$ and $\sec^{-1}(-x) = \pi - \sec^{-1}(x)$:

$\text{cosec}^{-1}(-x) + \sec^{-1}(-x)$

$= -\text{cosec}^{-1}(x) + \pi - \sec^{-1}(x)$

$= \pi - [\text{cosec}^{-1}(x) + \sec^{-1}(x)]$

Using $\text{cosec}^{-1}(x) + \sec^{-1}(x) = \frac{\pi}{2}$:

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

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

(B) matches with (IV)

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

The angle with tangent $\sqrt{3}$ is $\frac{\pi}{3}$

For $\sec^{-1}(-2)$, we need $\sec(\theta) = -2$, which means $\cos(\theta) = -\frac{1}{2}$

In the range $[0, \pi]$ excluding $\frac{\pi}{2}$, this gives $\theta = \frac{2\pi}{3}$

Therefore:

$\tan^{-1}\sqrt{3} - \sec^{-1}(-2)$

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

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

(C) matches with (II)

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For (D), since $\frac{4\pi}{3}$ is outside the range of $\tan^{-1}$, which is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$:

$\frac{4\pi}{3} = \pi + \frac{\pi}{3}$

Using $\tan(\pi + \theta) = \tan(\theta)$:

$\tan\frac{4\pi}{3} = \tan\frac{\pi}{3} = \sqrt{3}$

Therefore:

$\tan^{-1}\left(\tan\frac{4\pi}{3}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$

(D) matches with (I)

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Final Matching:

(A) → (III) = $\pi$

(B) → (IV) = $\frac{\pi}{2}$

(C) → (II) = $-\frac{\pi}{3}$

(D) → (I) = $\frac{\pi}{3}$