(a)[2]
Prove the identity by using trig identities to show that $\frac{1 - 2\sin^2\theta}{1 - \sin^2\theta} = 1 - \tan^2\theta$.
(b)[3]
Hence solve $\frac{1 - 2\sin^2\theta}{1 - \sin^2\theta} = 2\tan^2\theta$ for $0^\circ \leq \theta \leq 180^\circ$.