Proving $\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$ I got this question from a paper but can't solve it and the question paper has no solutions section.How do you prove this?
$$\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$$
Thanks in advance.

$$\mathrm{cosec} A+\mathrm{cotan} A=\frac{1}{\sin A}+\frac{\cos A}{\sin A}=\frac{1+\cos A}{\sin A}\\=\frac{1+\cos A}{\sin A}\frac{\cos A+\sin A-1}{\cos A+\sin A-1}\\=\frac{\cos A+\sin A-1+\cos^2 A+\sin A\cos A-\cos A}{\sin A\cos A+\sin^2 A-\sin A}\\\underbrace{=}_{(1)}\frac{\sin A-1+\cos^2 A+\sin A\cos A}{\sin A\cos A+\sin^2 A-\sin A}\\\underbrace{=}_{(2)}\frac{\sin A-\sin^2 A+\sin A\cos A}{\sin A\cos A+\sin^2 A-\sin A}\\\underbrace{=}_{(3)}\frac{1-\sin A+\cos A}{\cos A+\sin A-1},$$ where:
in $(1)$ we cancel $\cos A$ and $-\cos A$ in the numerator;
in $(2)$ we use $\sin^2A+\cos^2 A=1;$ 
in $(3)$ we divide numerator and denominator by $\sin A.$ 
