若sinB=m*sin(2A+B),其中M不为1,角A不等于90度的倍数,求证:tan(A+B)=(1+m)/(1-m)

sinB=m*sin(2A+B)
m=sinB/sin(2A+B)
1+m=[sin(2A+B)+sinB]/sin(2A+B)=2[sin(A+B)cosA]/sin(2A+B)
1-m=[sin(2A+B)-sinB]/sin(2A+B)=2[cos(A+B)sinA]/sin(2A+B)
(1+m)/(1-m)
=[sin(A+B)cosA]/[cos(A+B)sinA]
=tan(A+B)/tanA
tan(A+B)=(1+m)/(1-m)*tanA