(a) To prove the identity, start by using a common denominator:

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

Simplify the numerator:

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

Factor the numerator:

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

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

Thus, the expression becomes:

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

Therefore, the identity is proven.

(b) To solve the equation:

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

From part (a), we have:

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

Equating gives:

\(2 \tan^2 \theta = 0\)

Thus, \(\tan^2 \theta = 0\), leading to \(\tan \theta = 0\).

The solutions for \(\tan \theta = 0\) in the interval \(0 < \theta < 2\pi\) are \(\theta = \pi\).