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