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:

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