Question

xy+yz+zx=1 then the expression x+y/1-xy. +y+z/1-yz +z+x/1-zx is equal to

Answer 1

Please see the attachment

Answer 2

$Given \: \blue { xy+yz+zx = 1 \: ---(1) }$

$Now , \: \red{the \: value \: of }\\\red{ \frac{x+y}{1-xy} + \frac{y+z}{1-yz} + \frac{z+x}{1-zx}} \\= \frac{x+y}{xy+yz+zx-xy} + \frac{y+z}{xy+yz+zx-yz} + \frac{z+x}{xy+yz+zx-zx}$

$= \frac{x+y}{yz+zx} + \frac{y+z}{xy+zx} + \frac{z+x}{xy+yz}$

$= \frac{x+y}{z(x+y)} + \frac{y+z}{x(y+z)} + \frac{z+x}{y(x+z)}$

$= \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\= \frac{yz+xz+xy}{xyz} \\ = \frac{1}{xyz}$

Therefore.,

$Option \: \pink { (B) } \:is \: correct .$

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