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) \[\frac{-8}{[(1-{{x}^{2}})(1-{{y}^{2}})(1-{{z}^{2}})]}\]
Correct Answer: C
Solution :
\[x=\tan A,\,\,y=\tan B,\,\,-z=\tan C\]. 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 \[\Rightarrow \]\[\frac{2x}{1-{{x}^{2}}}+\frac{2y}{1-{{y}^{2}}}+\frac{2z}{1-{{z}^{2}}}\] \[=\frac{8xyz}{(1-{{x}^{2}})(1-{{y}^{2}})(1-{{z}^{2}})}\]
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