(i) Prove the identity \(\left( \frac{1}{\sin \theta} - \frac{1}{\tan \theta} \right)^2 \equiv \frac{1 - \cos \theta}{1 + \cos \theta}\).

(ii) Hence solve the equation \(\left( \frac{1}{\sin \theta} - \frac{1}{\tan \theta} \right)^2 = \frac{2}{5}\), for \(0^\circ \leq \theta \leq 360^\circ\).