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Prove the following\[\tan^{-1}\frac{2}{11}+\tan^{-1}\frac{7}{24}=\tan^{-1}\frac{1}{2}\]

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  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\Large \large\frac{x+y}{1-xy}\) \(\:if\:xy < 1\)
We have to prove $LHS: \tan^{-1}\large \large\frac{2}{11} +tan^{-1} \large \large\frac{7}{24}$$= RHS: tan^{-1}\large \frac{1}{2}$
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} +\large \frac{7}{24}$$ = \large \frac{24 \times 2}{24 \times 11} + \large\frac{11 \times 7}{11 \times 24}$$ $$= \large \frac{48+77}{264} =$ $\large\frac{125}{264}$
$ \Rightarrow 1 - xy = 1 - $$\large \frac{2}{11} \times \frac{7}{24}$$ = 1 - $$\large\frac{14}{264} = \large\frac{264 - 14}{264} = \large\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 $LHS: \tan^{-1}\large \frac{2}{11} +tan^{-1} \large \frac{7}{24}$$= tan^{-1}\large \frac{1}{2}$
=R.H.S.

 

answered Mar 2, 2013 by balaji.thirumalai
edited Mar 15, 2013 by thanvigandhi_1
 

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