To prove the identity \(\cot x - \cot 2x \equiv \csc 2x\), we start by using trigonometric identities.

Recall that \(\cot x = \frac{\cos x}{\sin x}\) and \(\cot 2x = \frac{\cos 2x}{\sin 2x}\).

Also, \(\csc 2x = \frac{1}{\sin 2x}\).

Using the double angle identities:

\(\sin 2x = 2 \sin x \cos x\)

\(\cos 2x = \cos^2 x - \sin^2 x\)

Now, substitute these into the expression:

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

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

Combine the fractions:

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

\(= \frac{2 \cos^2 x - \cos^2 x + \sin^2 x}{2 \sin x \cos x}\)

\(= \frac{\cos^2 x + \sin^2 x}{2 \sin x \cos x}\)

Using the Pythagorean identity \(\cos^2 x + \sin^2 x = 1\), we have:

\(= \frac{1}{2 \sin x \cos x}\)

\(= \frac{1}{\sin 2x}\)

\(= \csc 2x\)

Thus, the identity is proven: \(\cot x - \cot 2x \equiv \csc 2x\).