(i) Start with \(\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta}\).

Divide the numerator and the denominator by \(\cos \theta\):

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

This proves the identity.

(ii) Using the identity from part (i), \(\frac{\tan \theta - 1}{\tan \theta + 1} = \frac{\tan \theta}{6}\).

Let \(t = \tan \theta\), then \(\frac{t - 1}{t + 1} = \frac{t}{6}\).

Cross-multiply to get:

\(6(t - 1) = t(t + 1)\).

Expand and simplify:

\(6t - 6 = t^2 + t\).

Rearrange to form a quadratic equation:

\(t^2 - 5t + 6 = 0\).

Factorize the quadratic:

\((t - 2)(t - 3) = 0\).

So, \(t = 2\) or \(t = 3\).

Thus, \(\tan \theta = 2\) or \(\tan \theta = 3\).

Find \(\theta\) using inverse tangent:

\(\theta = \tan^{-1}(2) \approx 63.4^\circ\) or \(\theta = \tan^{-1}(3) \approx 71.6^\circ\).