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

Let’s start with the expression \[ \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} - \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}. \] **Step 1. Combine the two fractions using a common denominator.** The common denominator is \[ (\cos \theta - \sin \theta)(\cos \theta + \sin \theta) = \cos^2 \theta - \sin^2 \theta. \] Thus, we write the expression as \[ \frac{(\cos \theta + \sin \theta)^2 - (\cos \theta - \sin \theta)^2}{\cos^2 \theta - \sin^2 \theta}. \] **Step 2. Expand the numerators.** Evaluate: \[ (\cos \theta + \sin \theta)^2 = \cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta, \] \[ (\cos \theta - \sin \theta)^2 = \cos^2 \theta - 2\cos \theta \sin \theta + \sin^2 \theta. \] **Step 3. Subtract the two expanded squares.** Subtract to obtain: \[ (\cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta) - (\cos^2 \theta - 2\cos \theta \sin \theta + \sin^2 \theta) = 4\cos \theta \sin \theta. \] Now, the expression becomes \[ \frac{4\cos \theta \sin \theta}{\cos^2 \theta - \sin^2 \theta}. \] **Step 4. Express in terms of double angle formulas.** Recall the double angle identities: \[ \sin 2\theta = 2\sin \theta \cos \theta, \quad \cos 2\theta = \cos^2 \theta - \sin^2 \theta. \] Notice that the numerator \(4\cos \theta \sin \theta\) can be written as \[ 4\cos \theta \sin \theta = 2(2\sin \theta\cos \theta) = 2\sin 2\theta. \] Thus, the expression becomes \[ \frac{2\sin 2\theta}{\cos 2\theta}. \] **Step 5. Write the final result in terms of \(\tan 2\theta\).** Since \[ \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta}, \] we have \[ \frac{2\sin 2\theta}{\cos 2\theta} = 2\tan 2\theta. \] Thus, we have shown that \[ \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} - \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} = 2\tan 2\theta. \]