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# If $\tan^{-1}a+\tan^{-1}b+\tan^{-1}c=\pi,\;prove\;that\;a+b+c=abc.$

Toolbox:
• $tan^{-1}a+tan^{-1}b=tan^{-1}\large\frac{a+b}{1-ab}$
• $tan (\pi\: -\theta))=-tan \theta$
Given,$tan^{-1}a+tan^{-1}b+tan^{-1}c=\pi$
$tan^{-1}a+tan^{-1}b=\pi-tan^{-1}c$

Using the above formula of $tan^{-1}a+tan^{-1}b$ we get
$\Rightarrow tan^{-1} \large\frac{a+b}{1-ab}=\pi-tan^{-1}c$

take tan on both sides
$tan(tan^{-1} \large\frac{a+b}{1-ab})=tan(\pi- tan^{-1}c)$
taking $\theta=tan^{-1}c$ in the above formula we get
$\large\frac{a+b}{1-ab}=-tan\: tan^{-1}c$
$\large\frac{a+b}{1-ab}=-c$

$\Rightarrow\:a+b=-c+abc$
$\Rightarrow a+b+c=abc$

edited Mar 19, 2013