证明(1+secx+tanx)/(1+secx-tanx)=(1+sinx)/cosx
左边=（cosx/cosx+1/cosx+sinx/cosx)/（cosx/cosx+1/cosx-sinx/cosx)
=(cosx+1+sinx)/(cosx+1-sinx)
=(cosx+1+sinx)^2/[(cosx+1+sinx)(cosx+1-sinx)]
=(cos²x+sin²x+2*sinx*cosx+2*cosx+2*sinx+1)/(cos²x+2*cosx+1-sin²x)
=(cosx*sinx+cosx+sinx)/[(cosx+1)*cosx]
=(cosx+1)(sinx+1)/[(cosx+1)*cosx]
=（1＋sinx）／cosx
=右边
=(cos²x+sin²x+2*sinx*cosx+2*cosx+2*sinx+1)/(cos²x+2*cosx+1-sin²x)
=(cosx*sinx+cosx+sinx)/[(cosx+1)*cosx]
这个 上一步怎么转换到下一步的昂