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

To solve the equation \[ \frac { \cos \theta + \sin \theta } { \cos \theta - \sin \theta } - \frac { \cos \theta - \sin \theta } { \cos \theta + \sin \theta } = 2 \tan 2 \theta, \] let's simplify the left-hand side. 1. Combine the fractions on the left: \[ \frac{(\cos \theta + \sin \theta)^2 - (\cos \theta - \sin \theta)^2}{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta)}. \] 2. Use the difference of squares: The numerator can be expressed as: \[ [(\cos \theta + \sin \theta) - (\cos \theta - \sin \theta)][(\cos \theta + \sin \theta) + (\cos \theta - \sin \theta)]. \] This simplifies to: \[ (2 \sin \theta)(2 \cos \theta) = 4 \sin \theta \cos \theta. \] 3. The denominator is: \[ \cos^2 \theta - \sin^2 \theta = \cos 2\theta. \] 4. The complete expression now reads: \[ \frac{4 \sin \theta \cos \theta}{\cos 2\theta}. \] 5. Recognize that \(4 \sin \theta \cos \theta = 2 \sin 2\theta\): Thus, we can rewrite our equation as: \[ \frac{2 \sin 2\theta}{\cos 2\theta} = 2 \tan 2\theta. \] This shows us the left-hand side equals the right-hand side. Thus, the equation holds true, confirming: \[ \frac { \cos \theta + \sin \theta } { \cos \theta - \sin \theta } - \frac { \cos \theta - \sin \theta } { \cos \theta + \sin \theta } = 2 \tan 2 \theta. \] Therefore, the given equation is verified as true for all \(\theta\) where \(\cos \theta \neq \sin \theta\) and \(\cos 2\theta \neq 0\). If you have any specific ranges or conditions for \(\theta\) in mind, we can delve deeper into those aspects!