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If-xy-yz-zx-1-then-prove-that-x-1-x-2-y-1-y-2-z-1-z-2-4xyz-1-x-2-1-y-2-1-z-2-

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Question Number 109585 by nimnim last updated on 24/Aug/20
If (xy+yz+zx)=1, then prove that (x/(1−x^2 ))+(y/(1−y^2 ))+(z/(1−z^2 ))=((4xyz)/((1−x^2 )(1−y^2 )(1−z^2 )))
$${If}\:\left({xy}+{yz}+{zx}\right)=\mathrm{1},\:{then}\:{prove}\:{that} \\ $$$$\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{y}}{\mathrm{1}−{y}^{\mathrm{2}} }+\frac{{z}}{\mathrm{1}−{z}^{\mathrm{2}} }=\frac{\mathrm{4}{xyz}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right)\left(\mathrm{1}−{z}^{\mathrm{2}} \right)} \\ $$
Answered by som(math1967) last updated on 25/Aug/20
x(1−z^2 )(1−y^2 )+y(1−z^2 )(1−x^2 ) +z(1−x^2 )(1−y^2 ) =x+y+z−x^2 y−x^2 z−xyz −y^2 z−y^2 x−xyz−z^2 x−z^2 y−xyz +3xyz+xy^2 z+xyz^2 +x^2 yz =(x+y+z)−x(xy+yz+zx) −y(xy+yz+zx)−z(xy+yz+zx) +3xyz+xyz(xy+yz+zx) =x+y+z−x−y−z+4xyz ★ =4xyz ★∵xy+yz+zx=1 ∴(x/(1−x^2 ))+(y/(1−y^2 ))+(z/(1−z^2 )) =((4xyz)/((1−x^2 )(1−y^2 )(1−z^2 )))
$$\mathrm{x}\left(\mathrm{1}−\mathrm{z}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right)+\mathrm{y}\left(\mathrm{1}−\mathrm{z}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\:\:+\mathrm{z}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right) \\ $$$$=\mathrm{x}+\mathrm{y}+\mathrm{z}−\mathrm{x}^{\mathrm{2}} \mathrm{y}−\mathrm{x}^{\mathrm{2}} \mathrm{z}−\mathrm{xyz} \\ $$$$−\mathrm{y}^{\mathrm{2}} \mathrm{z}−\mathrm{y}^{\mathrm{2}} \mathrm{x}−\mathrm{xyz}−\mathrm{z}^{\mathrm{2}} \mathrm{x}−\mathrm{z}^{\mathrm{2}} \mathrm{y}−\mathrm{xyz} \\ $$$$+\mathrm{3xyz}+\mathrm{xy}^{\mathrm{2}} \mathrm{z}+\mathrm{xyz}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{yz} \\ $$$$=\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)−\mathrm{x}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right) \\ $$$$−\mathrm{y}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)−\mathrm{z}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right) \\ $$$$+\mathrm{3xyz}+\mathrm{xyz}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right) \\ $$$$=\mathrm{x}+\mathrm{y}+\mathrm{z}−\mathrm{x}−\mathrm{y}−\mathrm{z}+\mathrm{4xyz}\:\bigstar \\ $$$$=\mathrm{4xyz} \\ $$$$\bigstar\because\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=\mathrm{1} \\ $$$$\therefore\frac{\mathrm{x}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{y}}{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }+\frac{\mathrm{z}}{\mathrm{1}−\mathrm{z}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4xyz}}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{z}^{\mathrm{2}} \right)} \\ $$
Commented by nimnim last updated on 24/Aug/20
Thanks.
$${Thanks}. \\ $$
Commented by nimnim last updated on 24/Aug/20
Sir, I think the first line x(1−x^2 )(1−y^2 )+.. is typo mistake. Thank you once again.
$${Sir},\:{I}\:{think}\:{the}\:{first}\:{line} \\ $$$${x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right)+..\:{is}\:{typo}\:{mistake}. \\ $$$${Thank}\:{you}\:{once}\:{again}. \\ $$
Commented by som(math1967) last updated on 25/Aug/20
yes
$$\mathrm{yes}\: \\ $$

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