Question

已知2tanA=3tanB，求证：tan(A-B)=sin2B/5-cos2B

Answer 1

根据基本公式得到tan(A-B)=(tanA-tanB)/(1+tanAtanB)
然后将tanA=3/2tanB带入上式
得到tanB/[2+3（tanB）^2]
然后分子分母同乘以2（cosB）^2得到sin2B/[4（cosB）^2+6(sinB)^2]
再用公式（cosB）^2=1+2cosB和(sinB)^2=1-cosB
就可以得到sin2B/（5-cos2B）了

Answer 2

tan(A-B)=(tanA-tanB)/(1+tanAtanB)
=(tanB/2)/(1+3tan^2B/2)
=tanB/(2+3tan^2B)
=sinBcosB/(2cos^2B+3sin^2B)
=2sinBcosB/(4+2sin^2B)
=sin2B/(4+1-cos2B)
=sin2B/(5-cos2B)