1. **State the problem:** Prove the identity $$\frac{\cos^2 x - \sin^2 x}{\cos x - \sin x} = \cos x + \sin x.$$\n\n2. **Recall the formula:** The numerator is a difference of squares, which factors as $$a^2 - b^2 = (a-b)(a+b).$$ Here, $$\cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x).$$\n\n3. **Rewrite the expression using the factorization:**\n$$\frac{\cos^2 x - \sin^2 x}{\cos x - \sin x} = \frac{(\cos x - \sin x)(\cos x + \sin x)}{\cos x - \sin x}.$$\n\n4. **Cancel the common factor:** Since $$\cos x - \sin x \neq 0,$$ we can cancel it:\n$$\frac{\cancel{(\cos x - \sin x)}(\cos x + \sin x)}{\cancel{\cos x - \sin x}} = \cos x + \sin x.$$\n\n5. **Conclusion:** The original expression simplifies exactly to $$\cos x + \sin x,$$ proving the identity.