锐角a,b满足tan(a-b)=sin2b.证明2tan2b=tana+tanb.请给我一个比较完整清晰的解题,

Question

锐角a,b满足tan(a-b)=sin2b.证明2tan2b=tan a+tan b.
请给我一个比较完整清晰的解题,

Answer 1

tan(A-B)
= (tanA-tanB)/(1+tanAtanB)
∵tan(A-B)=sin(2B)=2sinBcosB
∴(tanA-tanB)/(1+tanAtanB) =2sinBcosB
∴tanA-tanB=2sinBcosB(1+tanAtanB)=2sinBcosB+2tanAsin²B
tanA-2tanAsin²B= 2sinBcosB+tanB
tanA(1-2sin²B)= 2sinBcosB+tanB
tanA=(2sinBcosB+tanB)/cos(2B)
=[sin(2B)+tanB]/cos(2B)
tanA+tanB
=[sin(2B)+tanB]/cos(2B)+tanB
={sin(2B)+tanB[1+cos(2B)]}/cos(2B)
= {sin(2B)+ sin(2B)}/cos(2B)
= 2tan(2B)