1求證(1+sinb-cosb)/(1+sinb+cosb)=tan(b/2)

熱心網友

證明：因為tan(b/2)=sin(b/2)/cos(b/2)=[2sin(b/2)cos(b/2)]/[2cos(b/2)cos(b/2)]=sinb/(1+cosb),tan(b/2)=sin(b/2)/cos(b/2)=[2sin(b/2)sin(b/2)]/[2sin(b/2)cos(b/2)]=(1-cosb)/sinb,所以，tan(b/2)=sinb/(1+cosb)=(1-cosb)/sinb=(1+sinb-cosb)/(1+sinb+cosb).(比例的等比性質)