在三角形ABC中,内角A,B,C的对边分别是a,b,c,且a2+b2+根号2ab=c2.

a2+b2+根号2ab=c2.
∴a^2+b^2-c^2=-√2ab
根据余弦定理：
cosC=(a^2+b^2-c^2)/(2ab)
=-√2ab/(2ab)
=-√2/2
∵C是三角形内角
∴C=135º（1）求角C（2）设COSACOSB=3根号2/5，(COS(α+A)COS(α+B))/COS²α=根号2/5，求TANα的值.∵cosAcosB=3√2/5∴1/2[cos(A+B)+cos(A-B)]=3√2/5∵A+B=π-C=π/4∴cos(A+B)=√2/2∴1/2[√2/2+cos(A-B)]=3√2/5∴cos(A-B)=7√2/10∵[cos(α+A)cos(α+B)]/cos²α=√2/5∴1/2[cos(2α+A+B)+cos(A-B)]/cos²α=√2/5∴ cos(2α+π/4)+7√2/10=2√2/5*cos²α∴ (cos2α-sin2α)√2/2+7√2/10=2√2/5*1/2(1+cos2α)∴3/10cos2α-1/2sin2α=-1/23/5*(1-tan²α)/(1+tan²α)-2tanα/(1+tan²α)=-13/5*(1-tan²α)-2tanα=-1-tan²αtan²α-5tanα+4=0tanα=1或tanα=4