1. **State the problem:** Prove the identity $$\frac{\cos A}{\sin A + \cos B} + \frac{\cos B}{\sin B - \cos A} = \frac{\cos A}{\sin A - \cos B} + \frac{\cos B}{\sin B + \cos A}$$ 2. **Rewrite the equation:** We want to show that the left-hand side (LHS) equals the right-hand side (RHS). 3. **Bring all terms to one side:** Consider $$\frac{\cos A}{\sin A + \cos B} + \frac{\cos B}{\sin B - \cos A} - \frac{\cos A}{\sin A - \cos B} - \frac{\cos B}{\sin B + \cos A} = 0$$ 4. **Group terms:** Group the terms with $\cos A$ and $\cos B$: $$\left(\frac{\cos A}{\sin A + \cos B} - \frac{\cos A}{\sin A - \cos B}\right) + \left(\frac{\cos B}{\sin B - \cos A} - \frac{\cos B}{\sin B + \cos A}\right) = 0$$ 5. **Factor out $\cos A$ and $\cos B$ respectively:** $$\cos A \left(\frac{1}{\sin A + \cos B} - \frac{1}{\sin A - \cos B}\right) + \cos B \left(\frac{1}{\sin B - \cos A} - \frac{1}{\sin B + \cos A}\right) = 0$$ 6. **Find common denominators and simplify each bracket:** For the first bracket: $$\frac{1}{\sin A + \cos B} - \frac{1}{\sin A - \cos B} = \frac{(\sin A - \cos B) - (\sin A + \cos B)}{(\sin A + \cos B)(\sin A - \cos B)} = \frac{-2 \cos B}{\sin^2 A - \cos^2 B}$$ For the second bracket: $$\frac{1}{\sin B - \cos A} - \frac{1}{\sin B + \cos A} = \frac{(\sin B + \cos A) - (\sin B - \cos A)}{(\sin B - \cos A)(\sin B + \cos A)} = \frac{2 \cos A}{\sin^2 B - \cos^2 A}$$ 7. **Substitute back:** $$\cos A \cdot \frac{-2 \cos B}{\sin^2 A - \cos^2 B} + \cos B \cdot \frac{2 \cos A}{\sin^2 B - \cos^2 A} = 0$$ Simplify: $$-2 \cos A \cos B \frac{1}{\sin^2 A - \cos^2 B} + 2 \cos A \cos B \frac{1}{\sin^2 B - \cos^2 A} = 0$$ 8. **Factor out $2 \cos A \cos B$:** $$2 \cos A \cos B \left(-\frac{1}{\sin^2 A - \cos^2 B} + \frac{1}{\sin^2 B - \cos^2 A}\right) = 0$$ 9. **Simplify the expression inside parentheses:** $$-\frac{1}{\sin^2 A - \cos^2 B} + \frac{1}{\sin^2 B - \cos^2 A} = \frac{\sin^2 B - \cos^2 A - (\sin^2 A - \cos^2 B)}{(\sin^2 A - \cos^2 B)(\sin^2 B - \cos^2 A)}$$ Simplify numerator: $$\sin^2 B - \cos^2 A - \sin^2 A + \cos^2 B = (\sin^2 B - \sin^2 A) + (\cos^2 B - \cos^2 A)$$ 10. **Use Pythagorean identities:** Recall $\sin^2 x + \cos^2 x = 1$, so $$\sin^2 B - \sin^2 A = (1 - \cos^2 B) - (1 - \cos^2 A) = -\cos^2 B + \cos^2 A$$ Therefore numerator becomes: $$(-\cos^2 B + \cos^2 A) + (\cos^2 B - \cos^2 A) = 0$$ 11. **Since numerator is zero, the whole fraction is zero:** $$2 \cos A \cos B \times 0 = 0$$ 12. **Conclusion:** The original expression equals zero, so the identity is proven. **Final answer:** The given trigonometric identity holds true.