已知1-tanA/1+tanA=2+根号3,则tan(π/4+A)是

已知1-tanA/1+tanA=2+根号3,则tan(π/4+A)是
已知1-tanA/1+tanA=2+根号3,则tan(π/4+A)是

已知1-tanA/1+tanA=2+根号3,则tan(π/4+A)是
tan(π/4+A)=sin(π/4+A)/cos(π/4+A)=(sinπ/4*cosA+cosπ/4*sinA)/(cosπ/4*cosA-sinπ/4*sinA)=(tanπ/4*cosA+sinA)/(cosA-tanπ/4*sinA)=(tanπ/4+tanA)/(1-tanπ/4*tanA)=(1+tanA)/(1-tanA)
因为1-tanA/1+tanA=2+根号3,所以(1+tanA)/(1-tanA)=1/(2+根号3)=2-根号3

tan(π/4+A)=sin(π/4+A)/cos(π/4+A)=(sinπ/4×cosA+cosπ/4*sinA)/(cosπ/4×cosA×sinπ/4×sinA)=(tanπ/4×cosA+sinA)/(cosA-tanπ/4*sinA)=(tanπ/4+tanA)/(1-tanπ/4*tanA)=(1+tanA)/(1-tanA)
∵1-tanA/1+tanA=2+√3，∴(1+tanA)/(1-tanA)=1/(2+根号3)=2-√3