1. **State the problem:** Prove that $$\frac{\tan \theta}{1+\sec \theta} + \frac{1+\sec \theta}{\tan \theta} = 2 \csc \theta$$. 2. **Recall definitions and identities:** - $\tan \theta = \frac{\sin \theta}{\cos \theta}$ - $\sec \theta = \frac{1}{\cos \theta}$ - $\csc \theta = \frac{1}{\sin \theta}$ 3. **Rewrite the left side using these identities:** $$\frac{\tan \theta}{1+\sec \theta} + \frac{1+\sec \theta}{\tan \theta} = \frac{\frac{\sin \theta}{\cos \theta}}{1 + \frac{1}{\cos \theta}} + \frac{1 + \frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}$$ 4. **Simplify the denominators:** $$1 + \frac{1}{\cos \theta} = \frac{\cos \theta + 1}{\cos \theta}$$ 5. **Substitute back:** $$\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta + 1}{\cos \theta}} + \frac{\frac{\cos \theta + 1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}$$ 6. **Simplify each fraction by multiplying numerator and denominator:** $$\frac{\sin \theta}{\cos \theta} \times \frac{\cos \theta}{\cos \theta + 1} + \frac{\cos \theta + 1}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}$$ 7. **Cancel common factors:** $$\frac{\sin \theta}{\cancel{\cos \theta}} \times \frac{\cancel{\cos \theta}}{\cos \theta + 1} + \frac{\cos \theta + 1}{\cancel{\cos \theta}} \times \frac{\cancel{\cos \theta}}{\sin \theta} = \frac{\sin \theta}{\cos \theta + 1} + \frac{\cos \theta + 1}{\sin \theta}$$ 8. **Find common denominator and combine:** $$\frac{\sin^2 \theta}{(\cos \theta + 1) \sin \theta} + \frac{(\cos \theta + 1)^2}{(\cos \theta + 1) \sin \theta} = \frac{\sin^2 \theta + (\cos \theta + 1)^2}{(\cos \theta + 1) \sin \theta}$$ 9. **Expand numerator:** $$(\cos \theta + 1)^2 = \cos^2 \theta + 2 \cos \theta + 1$$ 10. **Sum numerator terms:** $$\sin^2 \theta + \cos^2 \theta + 2 \cos \theta + 1 = 1 + 2 \cos \theta + 1 = 2 + 2 \cos \theta = 2(1 + \cos \theta)$$ 11. **Substitute back:** $$\frac{2(1 + \cos \theta)}{(\cos \theta + 1) \sin \theta}$$ 12. **Cancel $1 + \cos \theta$ terms:** $$\frac{2 \cancel{(1 + \cos \theta)}}{\cancel{(\cos \theta + 1)} \sin \theta} = \frac{2}{\sin \theta}$$ 13. **Recall that $\frac{1}{\sin \theta} = \csc \theta$, so:** $$\frac{2}{\sin \theta} = 2 \csc \theta$$ 14. **Conclusion:** The left side simplifies exactly to the right side, so the identity is proven. **Final answer:** $$\frac{\tan \theta}{1+\sec \theta} + \frac{1+\sec \theta}{\tan \theta} = 2 \csc \theta$$