Start with the left-hand side:

\(\frac{\sin^3 \theta}{\sin \theta - 1} - \frac{\sin^2 \theta}{1 + \sin \theta}\)

Find a common denominator:

\(\frac{\sin^3 \theta (1 + \sin \theta) - \sin^2 \theta (\sin \theta - 1)}{(\sin \theta - 1)(1 + \sin \theta)}\)

Simplify the numerator:

\(\sin^3 \theta + \sin^4 \theta - \sin^3 \theta + \sin^2 \theta = \sin^4 \theta + \sin^2 \theta\)

Thus, the expression becomes:

\(\frac{\sin^4 \theta + \sin^2 \theta}{(\sin \theta - 1)(1 + \sin \theta)}\)

Recognize that \(1 - \sin^2 \theta = \cos^2 \theta\), so:

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

Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have:

\(\tan^2 \theta (1 + \sin^2 \theta)\)

Therefore, the identity is proven as:

\(-\tan^2 \theta (1 + \sin^2 \theta)\)