(a)[1]
Verify the identity $(2x - 1)(4x^2 + 2x - 1) \equiv 8x^3 - 4x + 1$.
(b)[3]
Prove the identity $\dfrac{\tan^2 \theta + 1}{\tan^2 \theta - 1} \equiv \dfrac{1}{1 - 2\cos^2 \theta}$.
(c)[5]
Using the results of (a) and (b), solve the equation $\frac{\tan^2 \theta + 1}{\tan^2 \theta - 1} = 4\cos \theta$, for $0^\circ \leq \theta \leq 180^\circ$.