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.

To prove the identity \(\cos 4\theta - 4\cos 2\theta \equiv 8\sin^4 \theta - 3\), we start by using trigonometric identities.

1. Express \(\cos 4\theta\) using the double angle formula:

\(\cos 4\theta = 2\cos^2 2\theta - 1\)

2. Express \(\cos 2\theta\) using the double angle formula:

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

3. Substitute \(\cos 2\theta\) into \(\cos 4\theta\):

\(\cos 4\theta = 2(1 - 2\sin^2 \theta)^2 - 1\)

\(= 2(1 - 4\sin^2 \theta + 4\sin^4 \theta) - 1\)

\(= 2 - 8\sin^2 \theta + 8\sin^4 \theta - 1\)

\(= 8\sin^4 \theta - 8\sin^2 \theta + 1\)

4. Substitute into the left-hand side of the identity:

\(\cos 4\theta - 4\cos 2\theta = (8\sin^4 \theta - 8\sin^2 \theta + 1) - 4(1 - 2\sin^2 \theta)\)

\(= 8\sin^4 \theta - 8\sin^2 \theta + 1 - 4 + 8\sin^2 \theta\)

\(= 8\sin^4 \theta - 3\)

Thus, the identity is proven.

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.

To prove the identity \(\cos 4\theta + 4\cos 2\theta \equiv 8\cos^4\theta - 3\), we start by expressing \(\cos 4\theta\) and \(\cos 2\theta\) in terms of \(\cos \theta\).

First, use the double angle formula for \(\cos 2\theta\):

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

Next, express \(\cos 4\theta\) using the double angle formula again:

\(\cos 4\theta = \cos(2 \times 2\theta) = 2\cos^2(2\theta) - 1\)

Substitute \(\cos 2\theta = 2\cos^2\theta - 1\) into the expression for \(\cos 4\theta\):

\(\cos 4\theta = 2(2\cos^2\theta - 1)^2 - 1\)

Expand \((2\cos^2\theta - 1)^2\):

\((2\cos^2\theta - 1)^2 = 4\cos^4\theta - 4\cos^2\theta + 1\)

Substitute back:

\(\cos 4\theta = 2(4\cos^4\theta - 4\cos^2\theta + 1) - 1\)

\(\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 2 - 1\)

\(\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1\)

Now substitute \(\cos 4\theta\) and \(\cos 2\theta\) back into the original identity:

\(\cos 4\theta + 4\cos 2\theta = (8\cos^4\theta - 8\cos^2\theta + 1) + 4(2\cos^2\theta - 1)\)

\(= 8\cos^4\theta - 8\cos^2\theta + 1 + 8\cos^2\theta - 4\)

\(= 8\cos^4\theta - 3\)

Thus, the identity is proven: \(\cos 4\theta + 4\cos 2\theta \equiv 8\cos^4\theta - 3\).