(i) Start with the left-hand side:

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

Using \(\cos^2 \theta = 1 - \sin^2 \theta\), we have:

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

Factor the numerator:

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

Cancel \((1 - \sin \theta)\):

\(= \frac{1 + \sin \theta}{\sin \theta} = 1 + \frac{1}{\sin \theta}\)

Thus, the identity is proven.

(ii) Solve \(\frac{\cos \theta}{\tan \theta (1 - \sin \theta)} = 4\):

From part (i), \(1 + \frac{1}{\sin \theta} = 4\).

\(\frac{1}{\sin \theta} = 3\)

\(\sin \theta = \frac{1}{3}\)

Find \(\theta\) in the range \(0^\circ \leq \theta \leq 360^\circ\):

\(\theta = \arcsin \left( \frac{1}{3} \right) \approx 19.5^\circ\)

Also, \(\theta = 180^\circ - 19.5^\circ = 160.5^\circ\).

Thus, \(\theta = 19.5^\circ, 160.5^\circ\).