(i)[3]
Show that $\frac{\tan \theta + 1}{1 + \cos \theta} + \frac{\tan \theta - 1}{1 - \cos \theta} = \frac{2(\tan \theta - \cos \theta)}{\sin^2 \theta}$ is true.
(ii)[4]
Hence, showing all necessary working, solve the equation $\frac{\tan \theta + 1}{1 + \cos \theta} + \frac{\tan \theta - 1}{1 - \cos \theta} = 0$ within $0^\circ < \theta < 90^\circ$.