(a)[4]
Prove the identity $$\frac{\sin^3 \theta}{\sin \theta - 1} - \frac{\sin^2 \theta}{1 + \sin \theta} = -\tan^2 \theta (1 + \sin^2 \theta).$$
(b)[2]
Hence solve the equation $$\frac{\sin^3 \theta}{\sin \theta - 1} - \frac{\sin^2 \theta}{1 + \sin \theta} = \tan^2 \theta (1 - \sin^2 \theta)$$ within $0 < \theta < 2\pi$.