Starting with the left-hand side (LHS):

\(\frac{\tan^2 \theta + 1}{\tan^2 \theta - 1} = \frac{\frac{\sin^2 \theta}{\cos^2 \theta} + 1}{\frac{\sin^2 \theta}{\cos^2 \theta} - 1}\)

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

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

Using \(\sin^2 \theta + \cos^2 \theta = 1\):

\(= \frac{1}{1 - 2\cos^2 \theta}\)

This matches the right-hand side (RHS), thus proving the identity.