(a)[4]
Show that the expression $\dfrac{\sin \theta + 2\cos \theta}{\cos \theta - 2\sin \theta} - \dfrac{\sin \theta - 2\cos \theta}{\cos \theta + 2\sin \theta}$ is equal to $\dfrac{4}{5\cos^2 \theta - 4}$.
(b)[3]
Hence solve for $\theta$ in $0^\circ < \theta < 180^\circ$ the equation $\dfrac{\sin \theta + 2\cos \theta}{\cos \theta - 2\sin \theta} - \dfrac{\sin \theta - 2\cos \theta}{\cos \theta + 2\sin \theta} = 5$.