Question

If $x+y+z=xyz$, prove that $\frac{x+y}{1-xy}+\frac{y+z}{1-yz}+\frac{z+x}{1-zx}=\frac{x+y}{1-xy}\cdot \frac{y+z}{1-yz}\cdot \frac{z+x}{1-zx}$.

Answer 1

As above $A+B+C=0$ or $n\pi$
$\tan A+\tan B+\tan C=\tan A\tan B\tan C$ ..$\left(1\right)$
Now $A=\pi -(B+C)$
$\therefore \tan A=-\tan (B+C)$
or $\tan A=-\frac{\tan B+\tan C}{1-\tan B\tan C}=-\left(\frac{y+z}{1-yz}\right)$ etc.
Now put in $\left(1\right)$ and cancel $-$ive sign from both sides.

$\therefore \frac{x+y}{1-xy}+\frac{y+z}{1-yz}+\frac{z+x}{1-zx}=\frac{x+y}{1-xy}\cdot \frac{y+z}{1-yz}\cdot \frac{z+x}{1-zx}$.