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

\(\cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta}\)

Thus,

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

Combine the fractions:

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

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

\(\frac{1}{\sin \theta \cos \theta}\)

Now, express \(\csc 2\theta\):

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

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

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

Therefore,

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

Thus, \(\cot \theta + \tan \theta = 2 \csc 2\theta\), proving the identity.