Question

Let $S=\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}$ and
$T=\frac{8xyz}{(1-x^2)(1-y^2)(1-z^2)}$, where $x,y,z\ne \pm 1$. If $x+y+z=xyz$, then the value of $S-T$ is

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

Let $x=\tan A,y=\tan B,z=\tan C$
$S=\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}\Rightarrow S=\frac{2\tan A}{1-{\tan }^{2}A}+\frac{2\tan B}{1-{\tan }^{2}B}+\frac{2\tan C}{1-{\tan }^{2}C}\Rightarrow S=\tan 2A+\tan 2B+\tan 2C\cdots \left(1\right)$

$T=\frac{8xyz}{(1-x^2)(1-y^2)(1-z^2)}\Rightarrow T=\frac{2\tan A}{1-{\tan }^{2}A}\times \frac{2\tan B}{1-{\tan }^{2}B}\times \frac{2\tan C}{1-{\tan }^{2}C}\Rightarrow T=\tan 2A\tan 2B\tan 2C$

Given
$x+y+z=xyz\Rightarrow \tan A+\tan B+\tan C=\tan A\tan B\tan C$
So,
$\therefore A+B+C=n\pi ,n\in Z\Rightarrow 2A+2B+2C=2n\pi ,n\in Z\Rightarrow \tan (2A+2B+2C)=0\Rightarrow \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C\Rightarrow S=T\therefore S-T=0$