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

Recall that \(\cot \theta = \frac{\cos \theta}{\sin \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).

Thus, the left-hand side (LHS) becomes:

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

Now, use the double-angle identity for cosine: \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\).

So, the LHS becomes:

\(\frac{\cos 2\theta}{\sin \theta \cos \theta}.\)

Now, consider the right-hand side (RHS):

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

Using the double-angle identity for sine: \(\sin 2\theta = 2 \sin \theta \cos \theta\), we have:

\(2 \cdot \frac{\cos 2\theta}{2 \sin \theta \cos \theta} = \frac{\cos 2\theta}{\sin \theta \cos \theta}.\)

Thus, the RHS is equal to the LHS, proving the identity:

\(\cot \theta - \tan \theta \equiv 2 \cot 2\theta.\)

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.

To prove \(\frac{1 - \cos 2\theta}{1 + \cos 2\theta} \equiv \tan^2 \theta\), we start by using the double angle identity for cosine:

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

Substitute \(\cos 2\theta\) in the expression:

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

Simplify the expression:

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

Since \(1 - \sin^2 \theta = \cos^2 \theta\), we have:

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

Thus, \(\frac{1 - \cos 2\theta}{1 + \cos 2\theta} = \tan^2 \theta\) is proved.

To prove the identity \(\frac{2 \sin x - \sin 2x}{1 - \cos 2x} \equiv \frac{\sin x}{1 + \cos x}\), we start by simplifying the left-hand side (LHS).

First, use the double angle formula for sine: \(\sin 2x = 2 \sin x \cos x\).

Substitute \(\sin 2x\) in the LHS:

\(\frac{2 \sin x - 2 \sin x \cos x}{1 - \cos 2x}\)

Simplify the numerator:

\(2 \sin x (1 - \cos x)\)

Now, use the double angle formula for cosine: \(\cos 2x = 1 - 2 \sin^2 x\).

Substitute \(\cos 2x\) in the denominator:

\(1 - (1 - 2 \sin^2 x) = 2 \sin^2 x\)

Thus, the LHS becomes:

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

Cancel \(2 \sin x\) from the numerator and denominator:

\(\frac{1 - \cos x}{\sin x}\)

Multiply by \(\frac{\sin x}{\sin x}\) to rationalize:

\(\frac{\sin x (1 - \cos x)}{\sin x (1 + \cos x)} = \frac{\sin x}{1 + \cos x}\)

Thus, the identity is proven.

To prove the identity \(\frac{\cot x - \tan x}{\cot x + \tan x} \equiv \cos 2x\), we start by expressing \(\cot x\) and \(\tan x\) in terms of sine and cosine:

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

Substitute these into the left-hand side (LHS):

\(\frac{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}}{\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}}\)

Find a common denominator for the fractions:

\(\frac{\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}}{\frac{\cos^2 x + \sin^2 x}{\sin x \cos x}}\)

Simplify the expression:

\(\frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x}\)

Using the Pythagorean identity \(\cos^2 x + \sin^2 x = 1\), the expression simplifies to:

\(\cos^2 x - \sin^2 x\)

Recognize this as the double angle identity for cosine:

\(\cos 2x = \cos^2 x - \sin^2 x\)

Thus, the identity is proven:

\(\frac{\cot x - \tan x}{\cot x + \tan x} \equiv \cos 2x\)