Prove the identity: (sinx-cosx+1)/(sinx+cosx-1)=(sinx+1)/cosx?

L.H.S.
= (sinx - cosx + 1) / (sinx + cosx - 1)
= cosx (sinx - cosx + 1) / [cosx (sinx + cosx - 1)]
= (cosx sinx - cos²x + cosx) / [cosx (sinx + cosx - 1)]
= (cosx sinx - cos²x + cosx) / [cosx (sinx + cosx - 1)]
= [cosx sinx - (1 - sin²x) + cosx] / [cosx (sinx + cosx - 1)]
= [cosx sinx - 1 + sin²x + cosx] / [cosx (sinx + cosx - 1)]
= [(sin²x + cosx sinx - sinx) + (sinx + cosx - 1)] / [cosx (sinx + cosx - 1)]
= [sinx(sinx + cosx - 1) + (sinx + cosx - 1)] / [cosx (sinx + cosx - 1)]
= (sinx + 1) (sinx + cosx - 1) / [cosx (sinx + cosx - 1)]
= (sinx + 1) / cosx
= R.H.S.

Thus, (sinx - cosx + 1) / (sinx + cosx - 1) = (sinx + 1) / cosx
provided that (sinx + cosx - 1) ≠ 0 and cosx ≠ 0

　
(sinx−cosx+1) / (sinx+cosx−1)
= (sinx−cosx+1)(sinx+cosx+1) / ((sinx+cosx−1)(sinx+cosx+1))
= ((sinx+1)²−cos²x) / ((sinx+cosx)²−1)
= (sin²x + 2sinx + 1 − cos²x) / (sin²x + 2 sinx cosx + cos²x − 1)
= (sin²x + 2sinx + (1−cos²x)) / (2 sinx cosx + (sin²x+cos²x) − 1)
= (sin²x + 2sinx + sin²x) / (2 sinx cosx + 1 − 1)
= (2sin²x + 2sinx) / (2 sinx cosx)
= 2 sinx (sinx + 1) / (2 sinx cosx)
= (sinx + 1) / cosx