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Question Number 109585 by nimnim last updated on 24/Aug/20

$$If(xy+yz+zx)=1,thenprovethat$$$$\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}=\frac{4xyz}{(1-x^2)(1-y^2)(1-z^2)}$$

Answered by som(math1967) last updated on 25/Aug/20

$$\mathrm{x}(1-{\mathrm{z}}^2)(1-{\mathrm{y}}^2)+\mathrm{y}(1-{\mathrm{z}}^2)(1-{\mathrm{x}}^2)$$$$+\mathrm{z}(1-{\mathrm{x}}^2)(1-{\mathrm{y}}^2)$$$$=\mathrm{x}+\mathrm{y}+\mathrm{z}-{\mathrm{x}}^2\mathrm{y}-{\mathrm{x}}^2\mathrm{z}-\mathrm{xyz}$$$$-{\mathrm{y}}^2\mathrm{z}-{\mathrm{y}}^2\mathrm{x}-\mathrm{xyz}-{\mathrm{z}}^2\mathrm{x}-{\mathrm{z}}^2\mathrm{y}-\mathrm{xyz}$$$$+3\mathrm{xyz}+{\mathrm{xy}}^2\mathrm{z}+{\mathrm{xyz}}^2+{\mathrm{x}}^2\mathrm{yz}$$$$=(\mathrm{x}+\mathrm{y}+\mathrm{z})-\mathrm{x}(\mathrm{xy}+\mathrm{yz}+\mathrm{zx})$$$$-\mathrm{y}(\mathrm{xy}+\mathrm{yz}+\mathrm{zx})-\mathrm{z}(\mathrm{xy}+\mathrm{yz}+\mathrm{zx})$$$$+3\mathrm{xyz}+\mathrm{xyz}(\mathrm{xy}+\mathrm{yz}+\mathrm{zx})$$$$=\mathrm{x}+\mathrm{y}+\mathrm{z}-\mathrm{x}-\mathrm{y}-\mathrm{z}+4\mathrm{xyz}★$$$$=4\mathrm{xyz}$$$$★\because \mathrm{xy}+\mathrm{yz}+\mathrm{zx}=1$$$$\therefore \frac{\mathrm{x}}{1-{\mathrm{x}}^2}+\frac{\mathrm{y}}{1-{\mathrm{y}}^2}+\frac{\mathrm{z}}{1-{\mathrm{z}}^2}$$$$=\frac{4\mathrm{xyz}}{(1-{\mathrm{x}}^2)(1-{\mathrm{y}}^2)(1-{\mathrm{z}}^2)}$$

Commented by nimnim last updated on 24/Aug/20

$$Thanks.$$

Commented by nimnim last updated on 24/Aug/20

$$Sir,Ithinkthefirstline$$$$x(1-x^2)(1-y^2)+..istypomistake.$$$$Thankyouonceagain.$$

Commented by som(math1967) last updated on 25/Aug/20

$$\mathrm{yes}$$

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