(a) Start with the left-hand side:

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

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

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

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

Thus, the identity is proven.

(b) Given:

\(\frac{2}{\sin \theta \cos \theta} = \frac{6 \cos \theta}{\sin \theta}\)

\(\Rightarrow \frac{2}{\cos \theta} = 6\)

\(\Rightarrow \cos^2 \theta = \frac{1}{3}\)

\(\Rightarrow \cos \theta = \pm 0.5774\)

For \(0^\circ < \theta < 180^\circ\), \(\theta = 54.7^\circ, 125.3^\circ\).