Question

求证：sin2A/2cosA(1+tan*tanA/2)=tanA

Answer 1

∵tanA/2=sin(A/2)/cos(A/2)=[sin(A/2)·2sin(A/2)] / [cos(A/2)·2sin(A/2)]=(1-cosA)/sinA.

∴1+tanA·tanA/2=1+(sinA/cosA)·(1-cosA)/sinA=1+1/cosA-1=1/cosA

又sin2A/2cosA=2sinAcosA/2cosA=sinA.

∴sin2A/2cosA·(1+tanA·tanA/2)=sinA·1/cosA=sinA/cosA=tanA.