Question

If $x+y+z=xyz$, then the value of $x(1-y^2)(1-z^2)+y(1-z^2)(1-x^2)+z(1-x^2)(1-y^2)$ is

• A. $xyz$
• B. $2xyz$
• C. $4xyz$
• D. None of these

Answer 1

Answer: (C)

$4xyz$

Let $x=\tan A,y=\tan B$ and $z=\tan C$

Given $x+y+z=xyz$

$\Rightarrow \tan A+\tan B+\tan C=\tan A\tan B\tan C$

$\Rightarrow \tan A+\tan B=-\tan C+\tan A\tan B\tan C$

$=-\tan C(1-\tan A\tan B)\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C\Rightarrow \tan (A+B)=\tan (-C)$

$\Rightarrow A+B=n\pi -C\Rightarrow A+B+C=n\pi \Rightarrow 2A+2B+2C=2n\pi$

$\Rightarrow \tan (2A+2B+2C)=\tan 2n\pi =0$

$\Rightarrow \frac{[\tan 2A+\tan 2B+\tan 2C-\tan 2A\tan 2B\tan 2C]}{[1-\tan 2A\tan 2B-\tan 2B\tan 2C-\tan 2C\tan 2A]}=0$

$\Rightarrow \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C\Rightarrow \frac{2\tan A}{1-{\tan }^{2}A}.\frac{2\tan B}{1-{\tan }^{2}B}.\frac{2\tan C}{1-{\tan }^{2}C}$

$=\frac{2\tan A}{1-{\tan }^{2}A}.\frac{2\tan B}{1-{\tan }^{2}B}.\frac{2\tan C}{1-{\tan }^{2}C}$

Hence, on putting $\tan A=x,\tan B=y,\tan C=z$ we get

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

On multiplying both sides by $\frac{1}{2}(1-x^2)(1-y^2)(1-z^2),$ we get

$x(1-y^2)(1-z^2)+y(1-z^2)(1-x^2)+z(1-x^2)(1-y^2)=4xyz$