Question

If $x+y+z=xyz$ ,then the value of $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}$ is

• A. $\left(\frac{2x}{1+x^2}\right)\left(\frac{2y}{1+y^2}\right)\left(\frac{2z}{1+z^2}\right)$
• B. $\left(\frac{x}{1-x^2}\right)\left(\frac{y}{1-y^2}\right)\left(\frac{z}{1-z^2}\right)$
• C. $\left(\frac{2x}{1-x^2}\right)\left(\frac{2y}{1-y^2}\right)\left(\frac{2z}{1-z^2}\right)$
• D. $\left(\frac{x}{1+x^2}\right)\left(\frac{y}{1+y^2}\right)\left(\frac{z}{1+z^2}\right)$

Answer 1

The correct option is A $\left(\frac{2x}{1+x^2}\right)\left(\frac{2y}{1+y^2}\right)\left(\frac{2z}{1+z^2}\right)$

Given : $x+y+z=xyz$

Let us assume that, $x=\tan A,y=\tan B,z=\tan c$

Now, $x+y+z=xyz\Rightarrow \tan A+\tan B+\tan C=\tan A\tan B\tan C$
and we know that this expression holds true only when $A+B+C=\pi$

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

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

$\Rightarrow \tan (2A+2B+2C)=0[\because \tan 2\pi =0]$

$\Rightarrow \frac{\tan 2\text{A}+tan2\text{B}+\tan 2\text{C}-\tan 2\text{A}\tan 2\text{B}\tan 2\text{C}}{1-\tan 2\text{A}\tan 2\text{B}-\tan 2\text{B}\tan 2\text{C}-\tan 2\text{A}\tan 2\text{C}}=0[\because \tan (\text{A}+\text{B}+\text{C})=\frac{\tan \text{A}+\tan \text{B}+\tan \text{C}-\tan \text{A}\tan \text{B}\tan \text{C}}{1-\tan \text{A}\tan \text{B}-\tan \text{B}\tan \text{C}-\tan \text{A}\tan \text{C}}]$

$\Rightarrow \tan 2\text{A}+tan2\text{B}+\tan 2\text{C}-\tan 2\text{A}\tan 2\text{B}\tan 2\text{C}=0$

$\Rightarrow \tan 2\text{A}+tan2\text{B}+\tan 2\text{C}=\tan 2\text{A}\tan 2\text{B}\tan 2\text{C}$

$\Rightarrow \frac{2\tan A}{1-{\tan }^{2}A}+\frac{2\tan B}{1-{\tan }^{2}B}+\frac{2\tan C}{1-{\tan }^{2}C}=\left(\frac{2\tan A}{1-{\tan }^{2}A}\right)\left(\frac{2\tan B}{1-{\tan }^{2}B}\right)\left(\frac{2\tan C}{1-{\tan }^{2}C}\right)[\because \tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }]$

$\Rightarrow \frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\left(\frac{2x}{1-x^2}\right)\left(\frac{2y}{1-y^2}\right)\left(\frac{2z}{1-z^2}\right)$