Question

△ABC中,角ABC对边为abc,已知sinB/2=sinA/2*sinC/2求tanA/2*tanC/2

Answer 1

tanA/2 * tanC/2=(sinA/2*sinC/2)/(cosA/2*cosC/2)

=(sinB/2)/(cos((A+C)/2)+(sinA/2*sinC/2))

=(sinB/2)/(cos((A+C)/2)+(sinB/2))=(sinB/2)/(cos((π-B)/2)+(sinB/2))

=(sinB/2)/(sinB/2+sinB/2)=1/2