(i) Use the double angle formulas: \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\) and \(\sin 2\theta = 2\sin \theta \cos \theta\).

Substitute into \(f(\theta)\):

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

Simplify the expression:

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

\(f(\theta) = \frac{(\sin \theta + \cos \theta)^2}{(\cos \theta + \sin \theta)^2} = \tan \theta\).

(ii) The integral becomes:

\(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \tan \theta \, d\theta = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\sin \theta}{\cos \theta} \, d\theta\).

Let \(u = \cos \theta\), then \(du = -\sin \theta \, d\theta\).

The integral becomes:

\(-\int \frac{1}{u} \, du = -\ln |u|\).

Evaluate from \(\frac{\pi}{6}\) to \(\frac{\pi}{4}\):

\(-\ln \left| \cos \frac{\pi}{4} \right| + \ln \left| \cos \frac{\pi}{6} \right|\).

\(-\ln \frac{1}{\sqrt{2}} + \ln \frac{\sqrt{3}}{2} = \ln \frac{\sqrt{3}}{2} - \ln \frac{1}{\sqrt{2}}\).

\(= \ln \frac{\sqrt{3}}{2} + \ln \sqrt{2} = \ln \frac{\sqrt{3} \cdot \sqrt{2}}{2} = \ln \frac{\sqrt{6}}{2}\).

\(= \frac{1}{2} \ln \frac{3}{2}\).