(a) Show that \(\frac{\sin \theta}{1 - \sin \theta} - \frac{\sin \theta}{1 + \sin \theta} \equiv 2 \tan^2 \theta\).

(b) Hence solve the equation \(\frac{\sin \theta}{1 - \sin \theta} - \frac{\sin \theta}{1 + \sin \theta} = 8\), for \(0^\circ < \theta < 180^\circ\).