已知1-tanA/1+tanA=2+根号3,则tan(π/4+A)是
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![已知1-tanA/1+tanA=2+根号3,则tan(π/4+A)是](/uploads/image/z/608658-42-8.jpg?t=%E5%B7%B2%E7%9F%A51-tanA%2F1%2BtanA%3D2%2B%E6%A0%B9%E5%8F%B73%2C%E5%88%99tan%28%CF%80%2F4%2BA%29%E6%98%AF)
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