If $\log(a+b+c)=\log(a) + \log(b) + \log(c)$, prove that $$\log\left(\frac{2a}{1-a^2} +\frac{2b}{1-b^2} +\frac{2c}{1-c^2}\right) = \log\left(\frac{2a}{1-a^2}\right)+ \log\left(\frac{2b}{1-b^2}\right)+\log\left(\frac{2c}{1-c^2}\right) $$
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Assuming $a,b,c>0$, Let:
$a =\tan\alpha $
$b =\tan\beta $
$c =\tan\gamma$
Inserting it in the given relation of the question, we get:- $$\log(a+b+c)=\log(a) + \log(b) + \log(c)$$ $$=> a+b+c= abc => \tan\alpha +\tan\beta+\tan\gamma=\tan\alpha\tan\beta\tan\gamma$$ $$=> \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=-\tan\gamma=\tan(\pi-\gamma) $$
Taking $ \alpha ,\beta,\gamma $ to be in the first quadrant, we get $\alpha+\beta+\gamma=\pi => 2\alpha+2\beta+2\gamma=2\pi $
Notice that:
$\tan2\alpha+\tan2\beta+\tan2\gamma-\tan2\alpha\tan2\beta\tan2\gamma =0$
On transposition, taking log, and then subsequently substituting back our inital substitutions, we get: $$\log\left(\frac{2a}{1-a^2} +\frac{2b}{1-b^2} +\frac{2c}{1-c^2}\right) = \log\left(\frac{2a}{1-a^2}\right)+ \log\left(\frac{2b}{1-b^2}\right)+\log\left(\frac{2c}{1-c^2}\right)$$
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In fact, $\alpha+\beta=n\pi-\gamma$ where $n$ is any integer.Though it hardly affects the conclusion – lab bhattacharjee May 03 '15 at 09:23