已知sinB=msin(2x+B)且x+B≠kπ+π／2,（k∈z）,x≠kπ／2 （k∈z）,m≠1.求证：tan(x+B)=[(1+m)/(1-m)]tanx证明：∵sinB=msin(2x+B),∴m=sinB/sin(2x+B)故(1+m)/(1-m)=[1+sinB/sin(2x+B)]/[1-sinB/sin(2x+B)]=[sin(...