(i) Show that \(\frac{\tan \theta + 1}{1 + \cos \theta} + \frac{\tan \theta - 1}{1 - \cos \theta} \equiv \frac{2(\tan \theta - \cos \theta)}{\sin^2 \theta}\).

(ii) Hence, showing all necessary working, solve the equation \(\frac{\tan \theta + 1}{1 + \cos \theta} + \frac{\tan \theta - 1}{1 - \cos \theta} = 0\) for \(0^\circ < \theta < 90^\circ\).