Match List-I with List-II

Choose the correct answer from the options given below:

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

The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

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

Using identity:

$\cos^{-1} (-x) = \pi - \cos^{-1} x$

Therefore,

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

Hence, (A) matches with (III) $\pi$.

(B) $\csc^{-1} (-x) + \sec^{-1} (-x)$

Recall identities:

$\csc^{-1} (-x) = -\csc^{-1} x$

$\sec^{-1} (-x) = \pi - \sec^{-1} x$ (for $x > 1$)

Adding,

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

For particular $x$, this sum equals $\frac{\pi}{2}$.

Hence, (B) matches with (IV) $\frac{\pi}{2}$.

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

Calculate each:

$\tan^{-1} \sqrt{3} = \frac{\pi}{3}$

$\sec^{-1} (-2) = \pi - \sec^{-1} 2 = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

Difference:

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

Hence, (C) matches with (II) $-\frac{\pi}{3}$.

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

Since $\tan^{-1}$ principal value lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$, and

$\frac{4\pi}{3} = \pi + \frac{\pi}{3}$, tangent has period $\pi$ so

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

Therefore,

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

Hence, (D) matches with (I) $\frac{\pi}{3}$.