Start with the left-hand side: \(\frac{\sin x \tan x}{1 - \cos x}\).

Using \(\tan x = \frac{\sin x}{\cos x}\), rewrite as:

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

Use the identity \(\sin^2 x = 1 - \cos^2 x\):

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

Factor the numerator as a difference of squares:

\(\frac{(1 - \cos x)(1 + \cos x)}{\cos x (1 - \cos x)}\).

Cancel \(1 - \cos x\) from the numerator and denominator:

\(\frac{1 + \cos x}{\cos x} = \frac{1}{\cos x} + 1\).

Thus, the identity is proven: \(\frac{\sin x \tan x}{1 - \cos x} \equiv 1 + \frac{1}{\cos x}\).