证明：
(1+1/cosx+tanx)/(1+1/cosx-tanx)
=(cosx/cosx+1/cosx+sinx/cosx)/(cosx/cosx+1/cosx-sinx/cosx)
分子分母同时乘以cosx
=(cosx+1+sinx)/(cosx+1-sinx)
分子分母同时乘以cosx
=(cos²x+cosx+sinxcosx)/[cosx*(cosx+1-sinx)]
=(1-sin²x+cosx+sinxcosx)/[cosx*(cosx+1-sinx)]
=[(1-sinx)(1+sinx)+cosx(1+sinx)]/[cosx*(cosx+1-sinx)]
=(1+sinx)(1-sinx+cosx)/[cosx*(cosx+1-sinx)]
=(1+sinx)/cosx
∴ 等式成立.