1. **State the problem:** Prove or simplify the trigonometric identity $$\frac{1 + \csc x}{\sec x} = \cos x + \cot x$$. 2. **Recall definitions and formulas:** - $\csc x = \frac{1}{\sin x}$ - $\sec x = \frac{1}{\cos x}$ - $\cot x = \frac{\cos x}{\sin x}$ 3. **Rewrite the left-hand side (LHS) using these definitions:** $$\frac{1 + \csc x}{\sec x} = \frac{1 + \frac{1}{\sin x}}{\frac{1}{\cos x}}$$ 4. **Simplify the numerator:** $$1 + \frac{1}{\sin x} = \frac{\sin x}{\sin x} + \frac{1}{\sin x} = \frac{\sin x + 1}{\sin x}$$ 5. **Substitute back into the LHS:** $$\frac{\frac{\sin x + 1}{\sin x}}{\frac{1}{\cos x}} = \frac{\sin x + 1}{\sin x} \times \frac{\cos x}{1} = \frac{(\sin x + 1) \cos x}{\sin x}$$ 6. **Rewrite the right-hand side (RHS):** $$\cos x + \cot x = \cos x + \frac{\cos x}{\sin x} = \frac{\cos x \sin x}{\sin x} + \frac{\cos x}{\sin x} = \frac{\cos x \sin x + \cos x}{\sin x} = \frac{\cos x (\sin x + 1)}{\sin x}$$ 7. **Compare LHS and RHS:** Both sides equal $$\frac{(\sin x + 1) \cos x}{\sin x}$$, so the identity holds. **Final answer:** The identity $$\frac{1 + \csc x}{\sec x} = \cos x + \cot x$$ is true.