(a)[4]
Show that $\frac{\tan\theta}{1 + \cos\theta} + \frac{\tan\theta}{1 - \cos\theta}$ is equal to $\frac{2}{\sin\theta \, \cos\theta}$.
(b)[4]
Hence solve the equation $\dfrac{\tan \theta}{1 + \cos \theta} + \dfrac{\tan \theta}{1 - \cos \theta} = \dfrac{6}{\tan \theta}$ in the range $0^\circ < \theta < 180^\circ$.