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

To simplify the expression \[ \frac { \cos x + \sin x } { \cos x - \sin x } - \frac { \cos x - \sin x } { \cos x + \sin x } \] let's denote \( u = \cos x + \sin x \) and \( v = \cos x - \sin x \). The left-hand side (LHS) can be rewritten as follows: \[ \frac{u}{v} - \frac{v}{u} = \frac{u^2 - v^2}{uv} \] Now we compute \( u^2 \) and \( v^2 \): \[ u^2 = (\cos x + \sin x)^2 = \cos^2 x + 2\cos x \sin x + \sin^2 x = 1 + \sin(2x) \] \[ v^2 = (\cos x - \sin x)^2 = \cos^2 x - 2\cos x \sin x + \sin^2 x = 1 - \sin(2x) \] Thus: \[ u^2 - v^2 = (1 + \sin(2x)) - (1 - \sin(2x)) = 2\sin(2x) \] Next, we compute \( uv \): \[ uv = (\cos x + \sin x)(\cos x - \sin x) = \cos^2 x - \sin^2 x = \cos(2x) \] Substituting these back into our LHS: \[ \frac{u^2 - v^2}{uv} = \frac{2\sin(2x)}{\cos(2x)} = 2\tan(2x) \] Hence, we find that \[ \frac { \cos x + \sin x } { \cos x - \sin x } - \frac { \cos x - \sin x } { \cos x + \sin x } = 2 \tan 2x \] which shows that the original equation holds true.