Match List-I with List-II

List-I	List-II
(A) $\tan^{-1}\frac{2}{11} + \tan^{-1}\frac{7}{24}$	(I) $\frac{3\pi}{4}$
(B) $\tan^{-1}2 + \tan^{-1}3$	(II) $\pi$
(C) $\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3$	(III) $\tan^{-1}\frac{1}{2}$
(D) $\tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{13}$	(IV) $\tan^{-1}\frac{2}{9}$

Choose the correct answer from the options given below:

The key formula used here:

$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\dfrac{a+b}{1-ab}$, if $ab < 1$

$\tan^{-1}a + \tan^{-1}b = \pi + \tan^{-1}\dfrac{a+b}{1-ab}$, if $ab > 1$ and $a,b > 0$

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

Here $ab = \dfrac{2}{11} \times \dfrac{7}{24} = \dfrac{14}{264} < 1$

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

$= \tan^{-1}\left(\dfrac{\frac{48+77}{264}}{\frac{264-14}{264}}\right)$

$= \tan^{-1}\dfrac{125}{250}$

$= \tan^{-1}\dfrac{1}{2}$

$(A) \rightarrow (III)$

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

Here $ab = 2 \times 3 = 6 > 1$ and both positive, so $\pi$ is added.

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

$= \pi + \tan^{-1}\left(\dfrac{5}{-5}\right)$

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

$= \pi - \dfrac{\pi}{4}$

$= \dfrac{3\pi}{4}$

$(B) \rightarrow (I)$

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

From part (B), $\tan^{-1}2 + \tan^{-1}3 = \dfrac{3\pi}{4}$

$= \tan^{-1}1 + \dfrac{3\pi}{4}$

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

$= \pi$

$(C) \rightarrow (II)$

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(D) $\tan^{-1}\dfrac{1}{7} + \tan^{-1}\dfrac{1}{13}$

Here $ab = \dfrac{1}{7} \times \dfrac{1}{13} = \dfrac{1}{91} < 1$

$= \tan^{-1}\left(\dfrac{\frac{1}{7}+\frac{1}{13}}{1-\frac{1}{91}}\right)$

$= \tan^{-1}\left(\dfrac{\frac{13+7}{91}}{\frac{91-1}{91}}\right)$

$= \tan^{-1}\dfrac{20}{90}$

$= \tan^{-1}\dfrac{2}{9}$

$(D) \rightarrow (IV)$

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List-I	List-II
(A)	(III) $\tan^{-1}\frac{1}{2}$
(B)	(I) $\frac{3\pi}{4}$
(C)	(II) $\pi$
(D)	(IV) $\tan^{-1}\frac{2}{9}$