To prove \(\frac{1 - \cos 2\theta}{1 + \cos 2\theta} \equiv \tan^2 \theta\), we start by using the double angle identity for cosine:

\(\cos 2\theta = 1 - 2\sin^2 \theta\)

Substitute \(\cos 2\theta\) in the expression:

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

Simplify the expression:

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

Since \(1 - \sin^2 \theta = \cos^2 \theta\), we have:

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

Thus, \(\frac{1 - \cos 2\theta}{1 + \cos 2\theta} = \tan^2 \theta\) is proved.