To prove the identity \(\frac{2 \sin x - \sin 2x}{1 - \cos 2x} \equiv \frac{\sin x}{1 + \cos x}\), we start by simplifying the left-hand side (LHS).

First, use the double angle formula for sine: \(\sin 2x = 2 \sin x \cos x\).

Substitute \(\sin 2x\) in the LHS:

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

Simplify the numerator:

\(2 \sin x (1 - \cos x)\)

Now, use the double angle formula for cosine: \(\cos 2x = 1 - 2 \sin^2 x\).

Substitute \(\cos 2x\) in the denominator:

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

Thus, the LHS becomes:

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

Cancel \(2 \sin x\) from the numerator and denominator:

\(\frac{1 - \cos x}{\sin x}\)

Multiply by \(\frac{\sin x}{\sin x}\) to rationalize:

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

Thus, the identity is proven.