1. **Problem statement:** Given that $A+B+C=180^\circ$, prove that $$\cos^2 A + \cos^2 B - \cos^2 C = 1 - 2 \sin A \sin B \cos C.$$\n\n2. **Recall the identity:** Since $A+B+C=180^\circ$, we have $C = 180^\circ - (A+B)$. Also, use the cosine of a difference formula: $$\cos C = \cos(180^\circ - (A+B)) = -\cos(A+B).$$\n\n3. **Express $\cos^2 C$:** $$\cos^2 C = \cos^2(180^\circ - (A+B)) = \cos^2(A+B).$$\n\n4. **Expand $\cos^2(A+B)$:** Using the cosine addition formula, $$\cos(A+B) = \cos A \cos B - \sin A \sin B,$$ so $$\cos^2(A+B) = (\cos A \cos B - \sin A \sin B)^2 = \cos^2 A \cos^2 B - 2 \cos A \cos B \sin A \sin B + \sin^2 A \sin^2 B.$$\n\n5. **Substitute into the left side:** $$\cos^2 A + \cos^2 B - \cos^2 C = \cos^2 A + \cos^2 B - \cos^2(A+B) = \cos^2 A + \cos^2 B - \left(\cos^2 A \cos^2 B - 2 \cos A \cos B \sin A \sin B + \sin^2 A \sin^2 B\right).$$\n\n6. **Simplify:** $$= \cos^2 A + \cos^2 B - \cos^2 A \cos^2 B + 2 \cos A \cos B \sin A \sin B - \sin^2 A \sin^2 B.$$\n\n7. **Group terms:** $$= (\cos^2 A + \cos^2 B - \cos^2 A \cos^2 B) + 2 \cos A \cos B \sin A \sin B - \sin^2 A \sin^2 B.$$\n\n8. **Use Pythagorean identities:** Recall $\sin^2 x = 1 - \cos^2 x$, so $$- \sin^2 A \sin^2 B = - (1 - \cos^2 A)(1 - \cos^2 B) = - (1 - \cos^2 A - \cos^2 B + \cos^2 A \cos^2 B) = -1 + \cos^2 A + \cos^2 B - \cos^2 A \cos^2 B.$$\n\n9. **Substitute back:** $$= (\cos^2 A + \cos^2 B - \cos^2 A \cos^2 B) + 2 \cos A \cos B \sin A \sin B -1 + \cos^2 A + \cos^2 B - \cos^2 A \cos^2 B.$$\n\n10. **Combine like terms:** $$= 2(\cos^2 A + \cos^2 B - \cos^2 A \cos^2 B) + 2 \cos A \cos B \sin A \sin B -1.$$\n\n11. **Rewrite the expression:** Notice that $$2(\cos^2 A + \cos^2 B - \cos^2 A \cos^2 B) = 2(\cos^2 A (1 - \cos^2 B) + \cos^2 B) = 2(\cos^2 A \sin^2 B + \cos^2 B).$$\n\n12. **This is complicated; instead, use a simpler approach:**\nRecall the identity to prove: $$\cos^2 A + \cos^2 B - \cos^2 C = 1 - 2 \sin A \sin B \cos C.$$\n\n13. **Use the cosine double angle identity:** $$\cos^2 x = \frac{1 + \cos 2x}{2}.$$\n\n14. **Rewrite left side:** $$\cos^2 A + \cos^2 B - \cos^2 C = \frac{1 + \cos 2A}{2} + \frac{1 + \cos 2B}{2} - \frac{1 + \cos 2C}{2} = \frac{1}{2}(1 + \cos 2A + 1 + \cos 2B - 1 - \cos 2C) = \frac{1}{2}(1 + \cos 2A + \cos 2B - \cos 2C).$$\n\n15. **Rewrite right side:** $$1 - 2 \sin A \sin B \cos C.$$\n\n16. **Use product-to-sum formulas:** $$\sin A \sin B = \frac{\cos(A-B) - \cos(A+B)}{2}.$$\n\n17. **Substitute:** $$1 - 2 \sin A \sin B \cos C = 1 - 2 \cdot \frac{\cos(A-B) - \cos(A+B)}{2} \cos C = 1 - (\cos(A-B) - \cos(A+B)) \cos C.$$\n\n18. **Recall $C = 180^\circ - (A+B)$ and $\cos C = -\cos(A+B)$:**\nSo $$1 - (\cos(A-B) - \cos(A+B)) (-\cos(A+B)) = 1 + (\cos(A-B) - \cos(A+B)) \cos(A+B).$$\n\n19. **Expand:** $$= 1 + \cos(A-B) \cos(A+B) - \cos^2 (A+B).$$\n\n20. **Use product-to-sum for $\cos(A-B) \cos(A+B)$:** $$\cos(A-B) \cos(A+B) = \frac{\cos(2A) + \cos(2B)}{2}.$$\n\n21. **Substitute:** $$1 + \frac{\cos 2A + \cos 2B}{2} - \cos^2 (A+B).$$\n\n22. **Recall from step 3 that $\cos^2 C = \cos^2 (A+B)$, so the left side is:** $$\frac{1}{2}(1 + \cos 2A + \cos 2B - \cos 2C).$$\n\n23. **Since $\cos 2C = 2 \cos^2 C - 1$, rewrite:** $$\cos^2 C = \frac{1 + \cos 2C}{2}.$$\n\n24. **Therefore, the expressions on both sides match, confirming the identity.**\n\n**Final answer:** $$\boxed{\cos^2 A + \cos^2 B - \cos^2 C = 1 - 2 \sin A \sin B \cos C}.$$