A.math (1-secx)/(1+secx) + (1-cosx)/(1+cosx)=0

(1-secx)/(1+secx) + (1-cosx)/(1+cosx)

= [1 - 1/cosx] / [1 + 1/cosx] + [1 - cosx] / [1 + cosx]

= [(cosx - 1)/cosx] / [(cosx + 1)/cosx] + [1 - cosx] / [1 + cosx] 【通分母】

= [(cosx - 1)/cosx] [cosx/(cosx + 1)] + [1 - cosx] / [1 + cosx]

= (cosx - 1) / (cosx + 1) + (1 - cosx) / (cosx + 1)

= [(cosx - 1) + (1 - cosx)] / (cosx + 1)

= [cosx - 1 + 1 - cosx] / (cosx + 1)

= 0 / (cosx + 1)

= 0

Therefore (1-secx)/(1+secx) + (1-cosx)/(1+cosx)

2007-01-04 21:39:24 補充：
最後一行符號給吃掉了，應該為Therefore (1-secx)/(1+secx) + (1-cosx)/(1+cosx) = 0

(1-secx)/(1+secx) + (1-cosx)/(1+cosx)
(1 - 1/cos x)/(1 + 1/cos x) + (1 - cos x)/(1 + cos x)
= [(cos x - 1)/cos x] / [(cos x + 1)/cos x] + (1 - cos x)/(1 + cos x)
= -(1 - cos x)/(1 + cos x) + (1 - cos x)/(1 + cos x)
= 0

Making the same denominator,
LHS = [(1-secx+cosx-secxcosx)+(1-cosx+secx-cosxsecx)] / [(1+secx)(1+cosx)]
= (2 - 2cosxsecx) / [(1+secx)(1+cosx)]
= (2 - 2cosx/cosx) / [(1+secx)(1+cosx)]
= (2 - 2) / [(1+secx)(1+cosx)]
= 0
Since LHS = RHS,
Hence (1-secx)/(1+secx) + (1-cosx)/(1+cosx)=0.

Okay?

(1-secx)/(1+secx) + (1-cosx)/(1+cosx)
(1 - 1/cos x)/(1 + 1/cos x) + (1 - cos x)/(1 + cos x)
= [(cos x - 1)/cos x] / [(cos x + 1)/cos x] + (1 - cos x)/(1 + cos x)
= (cos x - 1)/(cos x + 1) + (1 - cos x)/(1 + cos x)
= -(1 - cos x)/(1 + cos x) + (1 - cos x)/(1 + cos x)
= 0

　LHS
= (1-secx)/(1+secx) + (1-cosx)/(1+cosx)
= (cosx-1)/(cosx+1) + (1-cosx)/(1+cosx)
= -(1-cosx)/(1+cosx) + (1-cosx)/(1+cosx)
= 0