Using trigonometry to prove $\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}=\frac{4abc}{\left(1-a^2\right)\left(1-b^2\right)\left(1-c^2\right)}.$ 
For the numbers $a,b,c$ with $ab+ac+bc=1$, prove that
$$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}=\frac{4abc}{\left(1-a^2\right)\left(1-b^2\right)\left(1-c^2\right)}.$$

This is a question from the trigonometric section of my textbook. But is it possible to use trigonometry here?
Edit:
Using the Dinesh Shankar's hint with $a=\tan\frac{x}{2}$, $b=\tan\frac{y}{2}$ and $c=\tan\frac{z}{2}$, the equation becomes
$$\tan x+\tan y+\tan z=\tan x\cdot\tan y\cdot\tan z.$$
But $z=\pi-(x+y)$, then
$$\tan x + \tan y + \tan \left(\pi-(x+y)\right)=\tan x\cdot\tan y\cdot\tan\left(\pi-(x+y)\right),$$
which implies
$$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\cdot\tan y}.$$
Therefore, the original equation is also valid.
Please feel free to give another solution.

Hint: 
You probably know that if $x+y+z=\pi$, then
$$\tan\frac{x}{2}\cdot\tan\frac{y}{2}+\tan\frac{x}{2}\cdot\tan\frac{z}{2}+\tan\frac{y}{2}\cdot\tan\frac{z}{2}=1.$$
Now, since $ab+ac+bc=1$, use $a=\tan\frac{x}{2}$....
I hope this help you.
