(i) Start with the left-hand side (LHS):

\(\frac{\sin \theta}{1 - \cos \theta} - \frac{1}{\sin \theta} = \frac{\sin^2 \theta - (1 - \cos \theta)}{(1 - \cos \theta) \sin \theta}\)

\(= \frac{1 - \cos^2 \theta - 1 + \cos \theta}{(1 - \cos \theta) \sin \theta}\)

\(= \frac{\cos \theta (1 - \cos \theta)}{(1 - \cos \theta) \sin \theta}\)

\(= \frac{\cos \theta}{\sin \theta} = \frac{1}{\tan \theta}\)

Thus, the identity is proven.

(ii) Solve \(\frac{1}{\tan \theta} = 4 \tan \theta\).

\(\tan^2 \theta = \frac{1}{4}\)

\(\tan \theta = \pm \frac{1}{2}\)

For \(\tan \theta = \frac{1}{2}\), \(\theta = 26.6^\circ\).

For \(\tan \theta = -\frac{1}{2}\), \(\theta = 153.4^\circ\).