1. **State the problem:** Prove the trigonometric identity $$\frac{\csc\theta}{\sin\theta} - \frac{\cot\theta}{\tan\theta} = 1$$. 2. **Recall definitions and formulas:** - $$\csc\theta = \frac{1}{\sin\theta}$$ - $$\cot\theta = \frac{\cos\theta}{\sin\theta}$$ - $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$ - Important Pythagorean identity: $$\csc^2\theta - \cot^2\theta = 1$$. 3. **Rewrite the left side using definitions:** $$\frac{\csc\theta}{\sin\theta} - \frac{\cot\theta}{\tan\theta} = \frac{\frac{1}{\sin\theta}}{\sin\theta} - \frac{\frac{\cos\theta}{\sin\theta}}{\frac{\sin\theta}{\cos\theta}}$$ 4. **Simplify each term:** $$\frac{1}{\sin\theta \cdot \sin\theta} - \frac{\frac{\cos\theta}{\sin\theta}}{\frac{\sin\theta}{\cos\theta}} = \frac{1}{\sin^2\theta} - \frac{\cos\theta}{\sin\theta} \cdot \frac{\cos\theta}{\sin\theta}$$ 5. **Simplify the second term:** $$\frac{\cos\theta}{\sin\theta} \cdot \frac{\cos\theta}{\sin\theta} = \frac{\cos^2\theta}{\sin^2\theta}$$ 6. **Rewrite the expression:** $$\frac{1}{\sin^2\theta} - \frac{\cos^2\theta}{\sin^2\theta} = \frac{1 - \cos^2\theta}{\sin^2\theta}$$ 7. **Use Pythagorean identity:** $$1 - \cos^2\theta = \sin^2\theta$$ 8. **Substitute back:** $$\frac{\sin^2\theta}{\sin^2\theta} = 1$$ 9. **Conclusion:** The left side simplifies to 1, so the identity is proven. **Final answer:** $$\frac{\csc\theta}{\sin\theta} - \frac{\cot\theta}{\tan\theta} = 1$$