Question

1+cos2a/{tana/2-(tana/2的负一次方)}化简

Answer 1

原式=(1+cos2A)/[tan(A/2)-1/tan(A/2)]=2cosA^2/[sin(A/2)/cos(A/2)-cos(A/2)/sin(A/2)]=2cosA^2*sin(A/2)*cos(A/2)/[sin(A/2)^2-cos(A/2)^2]=cosA^2*sinA/(-cosA)=-cosA*sinA=-sin(2A）/2 选择:A