(i) Start with the identity for \(\cot \theta\) and \(\tan \theta\):

\(\cot \theta = \frac{\cos \theta}{\sin \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).

Add them: \(\cot \theta + \tan \theta = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}\).

Using the double angle identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we have:

\(\frac{1}{\sin \theta \cos \theta} = \frac{2}{2 \sin \theta \cos \theta} = \frac{2}{\sin 2\theta} = 2 \csc 2\theta\).

Thus, \(\cot \theta + \tan \theta \equiv 2 \csc 2\theta\).

(ii) We need to evaluate \(\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} \csc 2\theta \, d\theta\).

Using the identity \(\csc 2\theta = \frac{1}{\sin 2\theta}\), the integral becomes:

\(\int \csc 2\theta \, d\theta = \int \frac{1}{\sin 2\theta} \, d\theta\).

Let \(u = 2\theta\), then \(du = 2 \, d\theta\) or \(d\theta = \frac{1}{2} \, du\).

The integral becomes \(\frac{1}{2} \int \csc u \, du\).

The integral of \(\csc u\) is \(\ln |\csc u - \cot u|\), so:

\(\frac{1}{2} \ln |\csc u - \cot u|\).

Substitute back \(u = 2\theta\):

\(\frac{1}{2} \ln |\csc 2\theta - \cot 2\theta|\).

Evaluate from \(\theta = \frac{1}{6}\pi\) to \(\theta = \frac{1}{3}\pi\):

\(\frac{1}{2} \left[ \ln |\csc \frac{2}{3}\pi - \cot \frac{2}{3}\pi| - \ln |\csc \frac{1}{3}\pi - \cot \frac{1}{3}\pi| \right]\).

Calculate the values:

\(\csc \frac{2}{3}\pi = \frac{1}{\sin \frac{2}{3}\pi} = \frac{2}{\sqrt{3}}\), \(\cot \frac{2}{3}\pi = \frac{1}{\tan \frac{2}{3}\pi} = -\frac{1}{\sqrt{3}}\).

\(\csc \frac{1}{3}\pi = \frac{1}{\sin \frac{1}{3}\pi} = 2\), \(\cot \frac{1}{3}\pi = \frac{1}{\tan \frac{1}{3}\pi} = \sqrt{3}\).

Substitute these back:

\(\frac{1}{2} \left[ \ln \left| \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} \right| - \ln \left| 2 - \sqrt{3} \right| \right]\).

\(\frac{1}{2} \left[ \ln \left( \frac{3}{\sqrt{3}} \right) - \ln \left( 2 - \sqrt{3} \right) \right]\).

\(\frac{1}{2} \left[ \ln 3 - \ln 1 \right] = \frac{1}{2} \ln 3\).