How to show that $\log (\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2})= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$ If $\log (a +b +c) =\log a+\log b+\log c$ then show that $$\log \left(\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2}\right)= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$$
Trial: If put $a=\frac{2a}{1-a^2}$ and the similar we are done. Can we do this in this way? Otherwise how to do. Please help. 

HINT:
Use  $\log a+\log b+\log c=\log(abc) $
and then put $a=\tan A$ etc. to find 
$\displaystyle \sum\tan A=\prod\tan A\implies A+B+C=n\pi$ where $n$ is any integer  (See here)
and $\displaystyle\frac{2a}{1-a^2}=\frac{2\tan A}{1-\tan^2A}=\tan2A$
$\displaystyle\implies \sum\tan2A=\prod\tan2A$  as $2A+2B+2C=2n\pi$
Now apply logarithm now
