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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

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    $\begingroup$ Put $x=\tan A,y=\tan B,z=\tan C$ $\endgroup$
    – DXT
    Feb 1, 2017 at 11:36
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    $\begingroup$ @DURGESHTIWARI You would have to prove that all such $x,y,z$ can be expressed in a form regarding $\tan A, \tan B, \tan C$ assuming $A+B+C=\pi$. It is known that if $x=\tan A, y=\tan B, z=\tan C$, then $$x+y+z=xyz$$ But you would ahve to proze the reverse holds. $\endgroup$
    – S.C.B.
    Feb 1, 2017 at 11:38
  • $\begingroup$ math-kali.blogspot.in/2010/07/if-x-y-z-xyz.html OR books.google.co.in/… $\endgroup$ Feb 1, 2017 at 15:59

2 Answers 2

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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.

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    $\begingroup$ @@Rohan, Are there other alternatives, too? $\endgroup$
    – pi-π
    Feb 1, 2017 at 12:04
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$\textbf{HINT:}$ 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.

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  • $\begingroup$ Note that $tan(\pi)=0$. $\endgroup$
    – att epl
    Feb 24, 2019 at 13:34

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