(i) Start with the expression:

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

Combine the fractions:

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

Simplify the numerator:

\(\tan \theta - \tan \theta \cos \theta + 1 - \cos \theta + \tan \theta + \tan \theta \cos \theta - 1 - \cos \theta\)

\(= 2\tan \theta - 2\cos \theta\)

The denominator is:

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

Thus, the expression simplifies to:

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

(ii) Set the expression to zero:

\(\frac{2(\tan \theta - \cos \theta)}{\sin^2 \theta} = 0\)

This implies:

\(\tan \theta - \cos \theta = 0\)

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

\(\sin \theta = \cos^2 \theta\)

\(\sin \theta = 1 - \sin^2 \theta\)

\(\sin^2 \theta + \sin \theta - 1 = 0\)

Using the quadratic formula, solve for \(\sin \theta\):

\(\sin \theta = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}\)

\(\sin \theta = 0.618\)

\(\theta = 38.2^\circ\)