Problem

(a) Prove the identity

$$\frac{\tan\theta + 7}{\tan^2\theta - 3} \equiv \frac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta}.$$

(b) Hence solve the equation

$$\frac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta} = \frac{5}{\tan\theta}$$

for $0^\circ \leq \theta \leq 180^\circ$.