設sinθ=cos^2θ，求1/(1-sinθ )+ 1/(1+sinθ)=?

Sol
Sinθ=Cos^2 θ>0
Sinθ=1－Sin^2 θ
Sin^2 θ+Sinθ－1=0
Sinθ=(－1+√5)/2 (負不合)
1/(1－Sinθ)+ 1/(1+Sinθ)
=[(1+Sinθ)+(1－Sinθ)]/[(1－Sinθ )(1+Sinθ)]
=2/Cos^2 θ
=2/Sinθ
=4/(－1+√5)
=(√5+1)

sinθ=cos^2θ
sinθ=1-sin^2θ
sin^2θ +sinθ-1=0
sinθ=(－1+√5)/2
因為(－1-√5)/2<0，所以√[(－1-√5)/2]不是實數的。

1/(1-sinθ )+ 1/(1+sinθ)
=[(1+sinθ)+(1-sinθ)]/[(1-sinθ)(1+sinθ)]
=2/(1-sin^2 θ)
=2/cos^2θ
=2/sinθ
=2/[(－1+√5)/2]
=4/(-1+√5)
=(4-4√5)/(-1-5)
=(2√5 -2)/3