1. State the problem: Prove that $$\frac{\cot \theta + \tan \theta}{\sec \theta} = \csc \theta.$$\n\n2. Write the trigonometric functions in terms of sine and cosine:\n$$\cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \sec \theta = \frac{1}{\cos \theta}.$$\n\n3. Substitute into the left-hand side (LHS):\n$$\frac{\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos \theta}}.$$\n\n4. Find common denominator for the numerator:\n$$\frac{\frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta}}{\frac{1}{\cos \theta}}.$$\n\n5. Recall the Pythagorean identity $$\cos^2 \theta + \sin^2 \theta = 1,$$ so numerator becomes:\n$$\frac{1}{\sin \theta \cos \theta}.$$\n\n6. The entire expression is now:\n$$\frac{\frac{1}{\sin \theta \cos \theta}}{\frac{1}{\cos \theta}} = \frac{1}{\sin \theta \cos \theta} \times \frac{\cos \theta}{1} = \frac{\cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta}.$$\n\n7. By definition, $$\csc \theta = \frac{1}{\sin \theta},$$ which matches the simplified expression.\n\nTherefore, the identity is proven: $$\frac{\cot \theta + \tan \theta}{\sec \theta} = \csc \theta.$$