(i) Start by combining the fractions on the left-hand side:

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

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

\(= \frac{4\cos \theta}{\sin^2 \theta}\)

\(= \frac{4}{\sin \theta \tan \theta}\)

Thus, the identity is proven.

(ii) Using the result from part (i), substitute into the equation:

\(\sin \theta \times \frac{4}{\sin \theta \tan \theta} = 3\)

\(\frac{4}{\tan \theta} = 3\)

\(\tan \theta = \frac{4}{3}\)

Solving for \(\theta\), we find:

\(\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.1^\circ\)

Considering the periodicity of the tangent function, the other solution in the range is:

\(\theta = 180^\circ + 53.1^\circ = 233.1^\circ\)