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

\[ \frac{\cos x}{\cos x - \sin x} - \frac{\cos x}{\cos x + \sin x} = \operatorname{Dtan} 2x \] First, find a common denominator: \[ \frac{\cos x (\cos x + \sin x) - \cos x (\cos x - \sin x)}{(\cos x - \sin x)(\cos x + \sin x)} = \operatorname{Dtan} 2x \] Simplify the numerator: \[ \cos^2 x + \cos x \sin x - \cos^2 x + \cos x \sin x = 2 \cos x \sin x \] So, the equation becomes: \[ \frac{2 \cos x \sin x}{\cos^2 x - \sin^2 x} = \operatorname{Dtan} 2x \] Recall that \(\cos^2 x - \sin^2 x = \cos 2x\) and \(2 \cos x \sin x = \sin 2x\): \[ \frac{\sin 2x}{\cos 2x} = \operatorname{Dtan} 2x \] Which simplifies to: \[ \tan 2x = \operatorname{Dtan} 2x \] Therefore, the equation holds true when: \[ \tan 2x = \operatorname{Dtan} 2x \] This implies that: \[ \operatorname{Dtan} 2x = \tan 2x \] So, the solution is: \[ \operatorname{Dtan} 2x = \tan 2x \]