Question

If G is the geometric mean of x and y, then
$t\frac{1}{G^2-x^2}+\frac{1}{G^2-y^2}=\frac{1}{G^2}$

Answer 1

$G^2=xy$

$\therefore$LHS=

$\frac{1}{G^2-x^2}+\frac{1}{G^2-y^2}$

$=\frac{1}{xy-x^2}+\frac{1}{xy-y^2}$

$=\frac{1}{x(y-x)}-\frac{1}{y(y-x)}$

$=\frac{1}{y-x}[\frac{1}{x}-\frac{1}{y}]$

$=\frac{1}{y-x}\left[\frac{y-x}{xy}\right]$

$=\frac{1}{xy}$

$=\frac{1}{G^2}=$RHS

Hence, proved