To prove the identity \(\csc 2\theta + \cot 2\theta \equiv \cot \theta\), we start by expressing \(\csc\) and \(\cot\) in terms of sine and cosine.

We know:

\(\csc 2\theta = \frac{1}{\sin 2\theta}, \quad \cot 2\theta = \frac{\cos 2\theta}{\sin 2\theta}\)

Using the double angle formulas:

\(\sin 2\theta = 2 \sin \theta \cos \theta, \quad \cos 2\theta = \cos^2 \theta - \sin^2 \theta\)

Substitute these into the identity:

\(\csc 2\theta + \cot 2\theta = \frac{1}{2 \sin \theta \cos \theta} + \frac{\cos^2 \theta - \sin^2 \theta}{2 \sin \theta \cos \theta}\)

Combine the fractions:

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

Since \(\cos^2 \theta - \sin^2 \theta = \cos 2\theta\), we have:

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

Using the identity \(1 + \cos 2\theta = 2 \cos^2 \theta\), we get:

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

Thus, the identity is proven.