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

问题描述:

已知2tanA=3tanB,求证tan(A-B)=

sin2B
5−cos2B

∵2tanA=3tanB,∴tan(A-B)=tanA−tanB1+tanAtanB=32tanB−tanB1+32tan2B=tanB2+3tan2B=sinBcosB2cos2B+3sin2B=2sinBcosB4cos2B+6sin2B=sin2B4+2sin2B=sin2B4+1−cos2B=sin2B5−cos2B.