Prove: If $x+y+z=xyz$ then $\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$

Question

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

Answer 1

Continuing from where you left, expressing terms of the numerator as: $$-xz^2-x^2z+x^2yz^2 =-xz (z+x-xyz) =-xz (-y) $$ $$-xy^2-x^2y+x^2y^2z =-xy (x+y-xyz) =-xy (-z) $$ $$-yz^2-y^2z+xy^2z^2 =-yz (y+z-xyz)=-yz (-x) $$ $$x+y+z=xyz $$

Now add everything and the result follows. Hope it helps.

Answer 2

$\textbf{HINT:}$ 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.