1. Stating the problem: Prove that $$\frac{1 - \sin \theta}{\cos \theta} = \frac{\cos \theta}{1 + \sin \theta}$$. 2. Start with the left-hand side (L.H.S): $$\frac{1 - \sin \theta}{\cos \theta}$$. 3. Multiply numerator and denominator by the conjugate of the numerator's denominator term, $$1 + \sin \theta$$, to rationalize the expression: $$\frac{1 - \sin \theta}{\cos \theta} \times \frac{1 + \sin \theta}{1 + \sin \theta} = \frac{(1 - \sin \theta)(1 + \sin \theta)}{\cos \theta (1 + \sin \theta)}$$. 4. Using the difference of squares formula, simplify the numerator: $$ (1)^2 - (\sin \theta)^2 = 1 - \sin^2 \theta $$. 5. Substitute this simplification back: $$\frac{1 - \sin^2 \theta}{\cos \theta (1 + \sin \theta)}$$. 6. Use the Pythagorean identity $$\sin^2 \theta + \cos^2 \theta = 1$$ to replace $$1 - \sin^2 \theta$$ with $$\cos^2 \theta$$: $$\frac{\cos^2 \theta}{\cos \theta (1 + \sin \theta)}$$. 7. Simplify by canceling one $$\cos \theta$$ from numerator and denominator: $$\frac{\cos \theta}{1 + \sin \theta}$$. 8. This matches the right-hand side (R.H.S), hence the identity is proved. Final answer: $$\frac{1 - \sin \theta}{\cos \theta} = \frac{\cos \theta}{1 + \sin \theta}$$.