Question

If $x+y+z=xyz$, prove that $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\cdot \frac{2y}{1-y^2}\cdot \frac{2z}{1-z^2}$.

Answer 1

Put $x=\tan A,y=\tan B$ and $z=\tan C$
Since $x+y+z=xyz$,
$\therefore \tan A+\tan B+\tan C-\tan A\tan B\tan C=0$
or $S_1-S_3=0$
Hence $\tan (A+B+C)=\frac{S_1-S_3}{1-S_2}=\frac{0}{1-S_2}=0$
$\therefore A+B=C=0$ or $n\pi$
or $2A+2B+2C=0$ or $2n\pi$
$\tan (2A+2B+2C)=0$
$\frac{S_1-S_3}{1-S_2}=0$ $\therefore S_1=S_3$
or $\tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$
Now put $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}=\frac{2x}{1-x^2}$ etc.
Another form:
Put $x=\tan A$ etc., then by given relation $x+y+z=xyz$
$\therefore \tan A+\tan B+\tan C=\tan A\tan B\tan C$.
Hence $A+B+C=\pi$ or $2A+2B+2C=2\pi$ etc.
$\therefore \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$.

$\therefore \frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\cdot \frac{2y}{1-y^2}\cdot \frac{2z}{1-z^2}$.