How can I prove that : (cosx +1)/sinx = (sinx-cosx-1)/ (1-cosx-sinx)

We'll cross multiply the terms of the
fractions:

sin x*(sin x - cos x - 1) = (cosx +1)(1 - cos x
- sin x)

We'll remove the
brackets:

(sin x)^2 - sin x*cos x - sin x = cos x - (cos
x)^2 - sin x*cos x + 1 - cos x - sin x

We'll eliminate sin
x*cos x both sides:

(sin x)^2  - sin x = cos x - (cos x)^2
+ 1 - cos x - sin x

We'll eliminate like terms both
sides:

 (sin x)^2 = - (cos x)^2 +
1

We'll add (cos x)^2 both
sides:

 (sin x)^2 + (cos x)^2 =
1

From the Pythagorean identity, we know that (sin x)^2 +
(cos x)^2 = 1.

The given identity is true:
(cosx +1)/sinx = (sinx-cosx-1)/ (1-cosx-sinx).