(tanA+secA-1)/(tanA-secA+1)=(1+sinA)/cosA
热心网友
(tanA+secA-1)/(tanA-secA+1)=(sinA+1-cosA)/(sinA-1+cosA)即只需证：(sinA+1-cosA)/(sinA-1+cosA)=(1+sinA)/cosA(sinA+1-cosA)*cosA=sinAcosA+cosA-cosA^2(sinA-1+cosA)*(1+sinA)=-1+cosA+sinA^2+sinAcosA=sinAcosA+cosA-cosA^2 (tanA+secA-1)/(tanA-secA+1)=(1+sinA)/cosA
热心网友
(tanA+secA-1)/(tanA-secA+1)=(sinA+1-cosA)/(sinA-1+cosA)=(2sinA/2cosA/2+2sin^2A/2)/(2sinA/2cosA/2-2sin^2A/2)=(cosA/2+sinA/2)/(cosA/2-sinA/2)=(cosA/2+sinA/2)^2/(cos^2A/2-sin^2A/2)=(1+2cosA/2sinA/2)/cosA=(1+sinA)/cosA