求證(tanθ+secθ-1)/(tanθ-secθ+1)=(1+sinθ)/cosθ
令(tanθ+secθ-1)/(tanθ-secθ+1)=k
1.利用合分比性質
(2tanθ)/2(secθ-1)=(k+1)/(k-1)
(sinθ/cosθ)/(1/cosθ)-1)=(1-cosθ)/sinθ
(1-cosθ)/(sinθ)=(k+1)/(k-1)
(1-cosθ)(1+cosθ)/(sinθ)(1+cosθ)=(k+1)/(k-1)
(sinθ)/(1+cosθ)=(k+1)/(k-1)
2.再利用合分比性質
(1+sinθ)/(cosθ)=(k+2)/(k-2)
故得證