To prove the identity \(\cot \theta - \tan \theta \equiv 2 \cot 2\theta\), we start by expressing each term in terms of sine and cosine.

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

Thus, the left-hand side (LHS) becomes:

\(\cot \theta - \tan \theta = \frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta}.\)

Now, use the double-angle identity for cosine: \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\).

So, the LHS becomes:

\(\frac{\cos 2\theta}{\sin \theta \cos \theta}.\)

Now, consider the right-hand side (RHS):

\(2 \cot 2\theta = 2 \cdot \frac{\cos 2\theta}{\sin 2\theta}.\)

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

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

Thus, the RHS is equal to the LHS, proving the identity:

\(\cot \theta - \tan \theta \equiv 2 \cot 2\theta.\)