Let \(t = \tan x = \frac{\sin x}{\cos x}\). Then \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\).

Substitute \(\tan^2 x\) into the identity:

\(\frac{1 - \frac{\sin^2 x}{\cos^2 x}}{1 + \frac{\sin^2 x}{\cos^2 x}}\)

Multiply numerator and denominator by \(\cos^2 x\):

\(\frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x}\)

Using the Pythagorean identity \(\cos^2 x + \sin^2 x = 1\), the expression simplifies to:

\(\cos^2 x - \sin^2 x\)

Using the identity \(\cos^2 x = 1 - \sin^2 x\), we have:

\((1 - \sin^2 x) - \sin^2 x = 1 - 2\sin^2 x\)

Thus, the identity is proven.