Prove: If $x+y+z=xyz$ then $\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$ If $x+y+z=xyz$, prove that:
$$\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$$.
My Attempt:
$$L.H.S=\frac {x}{1-x^2}+\frac {y}{1-y^2}+\frac {z}{1-z^2}$$
$$=\frac {x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)}{(1-x^2)(1-y^2)(1-z^2)}$$
$$=\frac {x+y+z-xz^2-xy^2+xy^2z^2-yz^2-yx^2+x^2yz^2-zy^2-zx^2+zx^2y^2}{(1-x^2)(1-y^2)(1-z^2)}$$.
I could not move on from here.Please help.
Thanks  

$\textbf{HINT:}$ Try putting $x=tan(\alpha);y=tan(\beta);z=tan(\gamma)$
Use: $$tan(2\theta)=\frac{2tan(\theta)}{1-tan^2(\theta)}$$
$\textbf{Note that $\alpha +\beta+\gamma=\pi$}$ by the condition since: $$tan(\alpha+\beta+\gamma)=\frac{\Sigma tan(\alpha)-tan(\alpha)tan(\beta)tan(\gamma)}{1-\Sigma tan(\alpha)tan(\beta)}$$ 
Where $\Sigma$ denotes cyclic summation.
