(i) To prove \(\cot x - \cot 2x \equiv \csc 2x\), use the identities:

\(\cot 2x = \frac{\cos 2x}{\sin 2x}\) and \(\csc 2x = \frac{1}{\sin 2x}\).

Express \(\cot x\) as \(\frac{\cos x}{\sin x}\).

Then, \(\cot x - \cot 2x = \frac{\cos x}{\sin x} - \frac{\cos 2x}{\sin 2x}\).

Using the double angle identity \(\cos 2x = 2\cos^2 x - 1\) and \(\sin 2x = 2\sin x \cos x\), simplify:

\(\cot x - \cot 2x = \frac{\cos x}{\sin x} - \frac{2\cos^2 x - 1}{2\sin x \cos x}\).

Combine the fractions:

\(\frac{2\cos^2 x - 1 - 2\cos^2 x}{2\sin x \cos x} = \frac{-1}{2\sin x \cos x} = \frac{-1}{\sin 2x} = -\csc 2x\).

Thus, \(\cot x - \cot 2x = \csc 2x\).

(ii) The indefinite integral of \(\cot x\) is \(\ln |\sin x|\).

Evaluate \(\int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \cot x \, dx = \left[ \ln |\sin x| \right]_{\frac{1}{6}\pi}^{\frac{1}{4}\pi}\).

Substitute the limits:

\(\ln |\sin \frac{1}{4}\pi| - \ln |\sin \frac{1}{6}\pi| = \ln \frac{\sin \frac{1}{4}\pi}{\sin \frac{1}{6}\pi}\).

\(\sin \frac{1}{4}\pi = \frac{\sqrt{2}}{2}\) and \(\sin \frac{1}{6}\pi = \frac{1}{2}\).

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

(iii) The indefinite integral of \(\csc 2x\) is \(\frac{1}{2} \ln |\tan x|\).

Evaluate \(\int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \csc 2x \, dx = \left[ \frac{1}{2} \ln |\tan x| \right]_{\frac{1}{6}\pi}^{\frac{1}{4}\pi}\).

Substitute the limits:

\(\frac{1}{2} \ln |\tan \frac{1}{4}\pi| - \frac{1}{2} \ln |\tan \frac{1}{6}\pi| = \frac{1}{2} \ln \frac{\tan \frac{1}{4}\pi}{\tan \frac{1}{6}\pi}\).

\(\tan \frac{1}{4}\pi = 1\) and \(\tan \frac{1}{6}\pi = \frac{1}{\sqrt{3}}\).

\(\frac{1}{2} \ln \frac{1}{\frac{1}{\sqrt{3}}} = \frac{1}{2} \ln \sqrt{3} = \frac{1}{4} \ln 3\).