$\displaystyle \frac{1-tan^2x}{ 1+tan^2x} = 1-2sin^2x$

$\displaystyle LHS = \frac{1 - \frac{sin^{2}x}{cos^{2}x}}{1+\frac{sin^{2}x}{cos^{ 2}x}}$

$\displaystyle =\frac{cos^{2}x-sin^{2}x}{1}$

$\displaystyle =1-sin^2x-sin^2x$

$\displaystyle =1-2sin^2x$

Am I on the right track?