• # question_answer If$x+y+z+xyz=0$, then$\frac{2x}{1-{{x}^{2}}}+\frac{2y}{1-{{y}^{2}}}-\frac{2z}{1-{{z}^{2}}}$ A) $\frac{xyz}{[(1-{{x}^{2}})(1-{{y}^{2}})(1-{{z}^{2}})]}$B) $\frac{-xyz}{[(1-{{x}^{2}})(1-{{y}^{2}})(1-{{z}^{2}})]}$C) $\frac{8xyz}{[(1-{{x}^{2}})(1-{{y}^{2}})(1-{{z}^{2}})]}$D) $\frac{-8}{[(1-{{x}^{2}})(1-{{y}^{2}})(1-{{z}^{2}})]}$

$x=\tan A,\,\,y=\tan B,\,\,-z=\tan C$. Then$(x+y-z)=-xyz$. $\Rightarrow$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$ $\Rightarrow$$A+B+C=\pi$         $\Rightarrow$$2A+2B=2\pi -2C$ $\Rightarrow$$\tan 2A+\tan 2B+\tan 2C=\tan 2A.\tan 2B.\tan 2C$$\Rightarrow$          $\frac{2\tan A}{1-{{\tan }^{2}}A}+\frac{2\tan B}{1-{{\tan }^{2}}B}+\frac{2\tan C}{1-{{\tan }^{2}}C}$ $=\frac{2\tan A}{1-{{\tan }^{2}}A}.\frac{2\tan B}{1-{{\tan }^{2}}B}.\frac{2\tan C}{1-{{\tan }^{2}}C}$ Put the value of $\tan A,\,\,\tan B,\,\,\tan C,$ we get $\Rightarrow$$\frac{2x}{1-{{x}^{2}}}+\frac{2y}{1-{{y}^{2}}}+\frac{2z}{1-{{z}^{2}}}$ $=\frac{8xyz}{(1-{{x}^{2}})(1-{{y}^{2}})(1-{{z}^{2}})}$