已知A,B,C是三角形ABC的三内角,向量m=(-1,根号3),n=(cosA,sinA),mn=1求:(1)A(2)若1+sin2B/cos^2B-sin^2B=-3,求tanB
问题描述：
已知A,B,C是三角形ABC的三内角,向量m=(-1,根号3),n=(cosA,sinA),mn=1
求:(1)A
(2)若1+sin2B/cos^2B-sin^2B=-3,求tanB
答
m*n=-cosA+√3sinA=2sin(A-π/6)=1 sin(A-π/6)=1/2 A-π/6=π/6A=π/3 (1+sin2B)/[(cosB)^2-(sinB)^2] =(sinB+cosB)^2/(sinB+cosB)(cosB-sinB)=(sinB+cosB)/(cosB-sinB)（分子分母同除cosB)=(tanB+1)/(1-tanB)=-3 t...