(i)[3]
Prove that the identity $\left( \frac{1}{\sin \theta} - \frac{1}{\tan \theta} \right)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$ holds.
(ii)[4]
Hence solve $\left( \frac{1}{ \sin \theta} - \frac{1}{\tan \theta} \right)^2 = \frac{2}{5}$, for $0^\circ \leq \theta \leq 360^\circ$.