Prove that (cos x - sin x + 1)/(cos x + sin x -1) = cosec x + cot x

You should use brackets appropriately. I have made the
relevant changes in your question.

We need to prove that
(cos x - sin x + 1)/(cos x + sin x -1) = cosec x + cot
x

Start from the left hand
side:

(cos x - sin x + 1)/(cos x + sin x
-1)

=> (cos x - sin x + 1)(cos x + sin x + 1)/(cos x
+ sin x - 1)(cos x + sin x + 1)

=> [(cos x + 1)^2 -
(sin x)^2]/[(cos x + sin x)^2 - 1]

=> [(cos x)^2 + 1
+ 2*cos x - (sin x)^2]/[(cos x)^2 + (sin x)^2 + 2*(sin x)(cos x) -
1]

=> [(cos x)^2 + 1 + 2*cos x - (sin x)^2]/[ 2*(sin
x)(cos x)]

=> [(cos x)^2 + (cos x)^2 + 2*cos
x]/2*(sin x)(cos x)

=> [2*(cos x)^2 + 2*cos
x]/2*(sin x)(cos x)

=> 2*(cos x)^2/2*(sin x)(cos x)
+ 2*cos x/2*(sin x)(cos x)

=> cos x/sin x + 1/cos
x

=> cot x + cosec
x

which is the right hand
side

This proves that (cos x - sin x +
1)/(cos x + sin x -1) = cosec x + cot x