(i) Start with the left-hand side: \(\frac{1 + \cos \theta}{\sin \theta} + \frac{\sin \theta}{1 + \cos \theta}\).

Combine the fractions: \(\frac{(1 + \cos \theta)^2 + \sin^2 \theta}{\sin \theta (1 + \cos \theta)}\).

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):

\((1 + \cos \theta)^2 + \sin^2 \theta = 1 + 2\cos \theta + \cos^2 \theta + 1 - \cos^2 \theta = 2 + 2\cos \theta\).

So, \(\frac{2 + 2\cos \theta}{\sin \theta (1 + \cos \theta)} = \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \frac{2}{\sin \theta}\).

This proves the identity.

(ii) Use the result from part (i): \(\frac{2}{\sin \theta} = \frac{3}{\cos \theta}\).

Cross-multiply: \(2\cos \theta = 3\sin \theta\).

Divide both sides by \(\cos \theta \sin \theta\): \(\frac{2}{\sin \theta} = \frac{3}{\cos \theta} \rightarrow t = \frac{2}{3}\).

\(\tan \theta = \frac{2}{3}\).

Find \(\theta\) using inverse tangent: \(\theta = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.7^\circ\).

Since \(\tan \theta\) is positive in the first and third quadrants, the solutions are \(\theta = 33.7^\circ\) and \(\theta = 180^\circ + 33.7^\circ = 213.7^\circ\).