Question

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

Answer 1

m*n=-cosA+√3sinA=2sin(A-π/6)=1
sin(A-π/6)=1/2
A-π/6=π/6
A=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
tanB=2