Match List-I with List-II List-I List-II (A) tan^-12/11+tan^-17/24 (I…

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

(A) $\tan^{-1}\left(\frac{2}{11}\right) + \tan^{-1}\left(\frac{7}{24}\right)$

Use identity: $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right)$ when $xy < 1$

Here, $x = \frac{2}{11},\ y = \frac{7}{24} \Rightarrow xy = \frac{14}{264} < 1$

$\Rightarrow \tan^{-1}\left(\frac{2}{11} + \frac{7}{24}\right) \div \left(1 - \frac{14}{264}\right)$

Numerator: $\frac{2}{11} + \frac{7}{24} = \frac{125}{264}$

Denominator: $1 - \frac{14}{264} = \frac{250}{264}$

$\Rightarrow \tan^{-1}\left(\frac{125}{264} \div \frac{250}{264}\right) = \tan^{-1}\left(\frac{1}{2}\right)$

So, (A) → (III)

(B) $\tan^{-1}(2) + \tan^{-1}(3)$

Here $xy = 6 > 1$, so use identity:

$\tan^{-1}(2) + \tan^{-1}(3) = \pi - \tan^{-1}\left(\frac{2 + 3}{1 - 6}\right) = \pi - \tan^{-1}(-1)$

$= \pi + \tan^{-1}(1) = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$

But it lies outside primary range, so in terms of sum: the effective identity gives $\frac{3\pi}{4}$

So, (B) → (I)

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

First add: $\tan^{-1}(1) + \tan^{-1}(2) = \tan^{-1}\left(\frac{3}{1 - 2} \right) = \tan^{-1}(-3)$

Now: $\tan^{-1}(-3) + \tan^{-1}(3) = 0$

Total: $0 + \pi = \pi$

So, (C) → (II)

(D) $\tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right)$

$\tan^{-1}\left(\frac{\frac{1}{7} + \frac{1}{13}}{1 - \frac{1}{91}}\right) = \tan^{-1}\left(\frac{\frac{20}{91}}{\frac{90}{91}}\right) = \tan^{-1}\left(\frac{2}{9}\right)$

So, (D) → (IV)