To prove the identity \(\tan 2\theta - \tan \theta \equiv \tan \theta \sec 2\theta\), we start by using the double angle formulas:

1. The double angle formula for tangent is:

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

2. The double angle formula for secant is:

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

3. Substitute \(\tan 2\theta\) into the left-hand side (LHS):

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

4. Combine into a single fraction:

\(\frac{2\tan \theta - \tan \theta (1 - \tan^2 \theta)}{1 - \tan^2 \theta} = \frac{2\tan \theta - \tan \theta + \tan^3 \theta}{1 - \tan^2 \theta}\)

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

5. Substitute \(\sec 2\theta\) into the right-hand side (RHS):

\(\tan \theta \sec 2\theta = \tan \theta \cdot \frac{2}{1 + \cos 2\theta}\)

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

6. Both sides are equal, thus proving the identity:

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

Therefore, the identity is verified.