To prove \(\csc 2\theta - \cot 2\theta \equiv \tan \theta\), we start by expressing the left-hand side in terms of \(\sin 2\theta\) and \(\cos 2\theta\).

We know:

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

Thus,

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

Using the double angle identities:

• \(\cos 2\theta = 1 - 2\sin^2 \theta\)
• \(\sin 2\theta = 2\sin \theta \cos \theta\)

Substitute these into the expression:

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

Simplify the fraction:

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

Thus, \(\csc 2\theta - \cot 2\theta \equiv \tan \theta\) is proven.