# 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
edited Mar 15, 2013