(a)[3]
Show that $\dfrac{\sin \theta}{1 - \sin \theta} - \dfrac{\sin \theta}{1 + \sin \theta} = 2 \tan^2 \theta$.
(b)[3]
Hence solve the equation $\dfrac{\sin \theta}{1 - \sin \theta} - \dfrac{\sin \theta}{1 + \sin \theta} = 8$, for $0^\circ < \theta < 180^\circ$.