1. **State the problem:** Verify the trigonometric identity: $$\frac{\sin\theta}{1-\cos\theta} + \frac{1-\cos\theta}{\sin\theta} = 2\csc\theta$$ 2. **Recall important formulas:** - $\csc\theta = \frac{1}{\sin\theta}$ - Use common denominator to combine fractions. 3. **Start with the left-hand side (LHS):** $$\frac{\sin\theta}{1-\cos\theta} + \frac{1-\cos\theta}{\sin\theta}$$ 4. **Find common denominator:** $$\frac{\sin^2\theta}{(1-\cos\theta)\sin\theta} + \frac{(1-\cos\theta)^2}{(1-\cos\theta)\sin\theta} = \frac{\sin^2\theta + (1-\cos\theta)^2}{(1-\cos\theta)\sin\theta}$$ 5. **Expand numerator:** $$(1-\cos\theta)^2 = 1 - 2\cos\theta + \cos^2\theta$$ So numerator is: $$\sin^2\theta + 1 - 2\cos\theta + \cos^2\theta$$ 6. **Use Pythagorean identity:** $$\sin^2\theta + \cos^2\theta = 1$$ So numerator becomes: $$1 + 1 - 2\cos\theta = 2 - 2\cos\theta = 2(1 - \cos\theta)$$ 7. **Substitute back:** $$\frac{2(1 - \cos\theta)}{(1-\cos\theta)\sin\theta}$$ 8. **Cancel common factor:** $$\frac{2\cancel{(1 - \cos\theta)}}{\cancel{(1-\cos\theta)}\sin\theta} = \frac{2}{\sin\theta}$$ 9. **Rewrite using cosecant:** $$\frac{2}{\sin\theta} = 2\csc\theta$$ 10. **Conclusion:** LHS equals RHS, so the identity is verified. **Final answer:** $$\frac{\sin\theta}{1-\cos\theta} + \frac{1-\cos\theta}{\sin\theta} = 2\csc\theta$$