Question

$Ifxy+yz+zx=1,$ prove that $\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}=\frac{4xyz}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}.$

Answer 1

Consider the given expression.

$\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}=\frac{4xyz}{(1-x^2)(1-y^2)(1-z^2)}$

$\frac{x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)}{(1-x^2)(1-y^2)(1-z^2)}=\frac{4xyz}{(1-x^2)(1-y^2)(1-z^2)}$

$x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)=4xyz$ …… (1)

L.H.S

$=x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)$

$=x(1-z^2-y^2+y^2{z}^2)+y(1-x^2-z^2+x^2{z}^2)+z(1-y^2-x^2+y^2{x}^2)$

$=x-x{z}^2-x{y}^2+x{y}^2{z}^2+y-y{x}^2-y{z}^2+y{x}^2{z}^2+z-z{y}^2-z{x}^2+z{y}^2{x}^2$

$=x+y+z-x{z}^2-x{y}^2-y{x}^2-y{z}^2-z{y}^2-z{x}^2+z{y}^2{x}^2+x{y}^2{z}^2+y{x}^2{z}^2$

$=x+y+z-(x{z}^2+x{y}^2+y{x}^2+y{z}^2+z{y}^2+z{x}^2)+xyz(xy+yz+xz)$

$=x+y+z-(x(xy+xz)+y(xy+yz)+z(yz+zx))+xyz(xy+yz+xz)$

Since, $xy+yz+zx=1$

Therefore,

$=x+y+z-(x(1-yz)+y(1-zx)+z(1-xy))+xyz$

$=x+y+z-(x-xyz+y-xyz+z-xyz)+xyz$

$=x+y+z-x+xyz-y+xyz-z+xyz+xyz$

$=4xyz$

Hence, proved.