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What is the value of $\tan^{-1}\large \frac{2}{11}$$ +tan^{-1} \large \frac{7}{24}$?

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  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\Large \frac{x+y}{1-xy}\) \(\:if\:xy < 1\)
Given $\tan^{-1}\large \frac{2}{11}$$ +tan^{-1} \large \frac{7}{24}$
Let $ x = \large \frac{2}{11}$ and $y = \large \frac{7}{24}$.
We know that \( tan^{-1}x+tan^{-1}y=tan^{-1}\Large \frac{x+y}{1-xy}\) \(\:if\:xy < 1\)
$ \Rightarrow x+y = \large \frac{2}{11} + \frac{7}{24}$$ = \large \frac{24 \times 2}{24 \times 11} + \frac{11 \times 7}{11 \times 24}$$ $$= \large \frac{48+77}{264} =$ $\large\frac{125}{264}$
$ \Rightarrow 1 - xy = 1 - $$\large \frac{2}{11} \times \frac{7}{24}$$ = 1 - $$\large\frac{14}{264} = \frac{264 - 14}{264} = \frac{250}{264}$
$\Rightarrow \large\frac{x+y}{1-xy}$$ = $$\large\frac{125}{264}$$ \times $${\large\frac{264}{250}}$$ = $$\large\frac{125}{250}$ $=$$\large \frac{1}{2}$
Substituting $\large\frac{x+y}{1-xy} = \frac{1}{2}$, we get $\tan^{-1}\large \frac{2}{11} +$$tan^{-1} \large \frac{7}{24}$ = $tan^{-1}\large \frac{1}{2}$
answered Mar 14, 2013 by balaji.thirumalai
 

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