<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] $$2x(1-y^2)(1-z^2)=2x-2xy^2-2xz^2+2xy^2z^2$$
$$2xy^2z^2=2yz(x+y+z)=2xyz + 2y^2z+2yz^2$$
$$\implies 2x(1-y^2)(1-z^2)=2x-2xy^2-2xz^2+2xyz+2y^2z+2yz^2$$
$$\implies 2y(1-z^2)(1-x^2)=2y-2yz^2-2yx^2+2xyz+2z^2x+2zx^2$$
$$\implies 2z(1-x^2)(1-y^2)=2z-2zx^2-2zy^2+2xyz+2x^2y+2xy^2$$
Summing up gives $2(x+y+z+3xyz)=8xyz$, which reduces to the equality.
I wish to see a simpler proof as well.
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