<s>[INST] If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$ 
If $x+y+z=xyz$, prove
$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
given that $x^2~,~y^2~,~z^2\ne1$

I came across this question in an ancient ($19$th century) Trigonometry book, and this is the method they use to prove the result (please note: I understand this method fully):
Let $x=\tan A$ ,$~y=\tan B$ and $z=\tan C$ which is acceptable without any loss of generality.
This means that we are saying that
$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$
Consider $\tan (A+B+C)$:
$$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1- \tan A \tan B- \tan C \tan A - \tan B \tan C}$$
So if $~\tan A+\tan B+\tan C=\tan A\tan B\tan C~~$ then $~~\tan (A+B+C)=0$. Hence, let $~A+B+C=\pi$.
Now consider $\tan (2A+2B+2C)$:
$$\tan (2A+2B+2C)=\frac{\tan 2A+\tan 2B+\tan 2C-\tan 2A\tan 2B\tan 2C}{1- \tan 2A \tan 2B- \tan 2C \tan 2A - \tan 2B \tan 2C}=0$$
$$\implies \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$$
$$\implies \frac{2\tan A}{1-\tan^2 A}+\frac{2\tan B}{1-\tan^2 B}+\frac{2\tan C}{1-\tan^2 C}=\frac{2\tan A}{1-\tan^2 A}\times\frac{2\tan B}{1-\tan^2 B}\times\frac{2\tan C}{1-\tan^2 C}$$
$$\therefore\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$$
as required.

My question is, is there any other way of proving this without this rather heavy use of trigonometry? I also would prefer not to work through heaps of algabraic manipulation and expansion to obtain the required result, although if that's necessary I will put up with it ;)
Thank you for your help.
 [/INST] Multiplying the eq. to prove by $\frac{(1-x^2)(1-y^2)(1-z^2)}{2}$ you obtain :
$$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2} \iff$$
$$x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)= 4xyz$$
but expanding the left side under the hypothesis $x+y+z=xyz\ $, you get :
$$x y^2 z^2 - x y^2 - x z^2 + x +x^2 y z^2 - x^2 y - y z^2 + y+ x^2 y^2 z - x^2 z - y^2 z + z =$$
$$x+y+z+xyz(yz+xy+xz)- x y^2 - x z^2- x^2 y - y z^2 - x^2 z - y^2 z=$$
$$(x+y+z)(yz+xy+xz+1)- x y^2 - x z^2- x^2 y - y z^2 - x^2 z - y^2 z=$$
$$xyz+xyz+xyz+x+y+z=3xyz+x+y+z = 4xyz$$
</s>