1. **State the problem:** Prove the identity $$\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta$$ 2. **Rewrite all trigonometric functions in terms of $\sin \theta$ and $\cos \theta$:** Recall: $$\tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \sec \theta = \frac{1}{\cos \theta}, \quad \csc \theta = \frac{1}{\sin \theta}$$ 3. **Rewrite the left-hand side (LHS):** $$\frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}$$ 4. **Simplify the denominators:** $$1 - \frac{\cos \theta}{\sin \theta} = \frac{\sin \theta - \cos \theta}{\sin \theta}$$ $$1 - \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta - \sin \theta}{\cos \theta}$$ 5. **Rewrite each fraction:** $$\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} = \frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta - \cos \theta} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)}$$ $$\frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}} = \frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta - \sin \theta} = \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$$ 6. **Note that $\cos \theta - \sin \theta = - (\sin \theta - \cos \theta)$, so rewrite the second term denominator:** $$\frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)} = \frac{\cos^2 \theta}{\sin \theta (- (\sin \theta - \cos \theta))} = - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$$ 7. **Sum the two terms:** $$\frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)} = \frac{\sin^2 \theta \sin \theta - \cos^2 \theta \cos \theta}{\cos \theta \sin \theta (\sin \theta - \cos \theta)}$$ 8. **Rewrite numerator:** $$\sin^3 \theta - \cos^3 \theta$$ 9. **Factor numerator as difference of cubes:** $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ So, $$\sin^3 \theta - \cos^3 \theta = (\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)$$ 10. **Use Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$:** $$\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta = 1 + \sin \theta \cos \theta$$ 11. **Substitute back:** $$\frac{(\sin \theta - \cos \theta)(1 + \sin \theta \cos \theta)}{\cos \theta \sin \theta (\sin \theta - \cos \theta)}$$ 12. **Cancel $\sin \theta - \cos \theta$ terms:** $$\frac{\cancel{(\sin \theta - \cos \theta)} (1 + \sin \theta \cos \theta)}{\cos \theta \sin \theta \cancel{(\sin \theta - \cos \theta)}} = \frac{1 + \sin \theta \cos \theta}{\cos \theta \sin \theta}$$ 13. **Split the fraction:** $$\frac{1}{\cos \theta \sin \theta} + \frac{\sin \theta \cos \theta}{\cos \theta \sin \theta} = \frac{1}{\cos \theta \sin \theta} + 1$$ 14. **Rewrite the first term using sec and csc:** $$\frac{1}{\cos \theta \sin \theta} = \frac{1}{\cos \theta} \times \frac{1}{\sin \theta} = \sec \theta \csc \theta$$ 15. **Therefore, the left-hand side equals:** $$1 + \sec \theta \csc \theta$$ which is the right-hand side. **Hence, the identity is proven.**