若cosA-2sin(A-B)=0,求证:tan(A-B)=cosB/(sinB+2)

问题描述:

若cosA-2sin(A-B)=0,求证:tan(A-B)=cosB/(sinB+2)

cosA-2sin(A-B)=cos[A-B+B]-2sin(A-B)=cos[(A-B)+B]-2sin(A-B)
=cos(A-B)cosB-sin(A-B)sinB-2sin(A-B)=0
=> cos(A-B)cosB=sin(A-B)[sinB+2]
=> tan(A-B)=cosB/(sinB+2)