cosA /1-tanA +sinA / 1-cotA =sinA+cosA

First off, work on the complicated side, which is the LHS.

Usually if we see tan A and cot A in an expression together with sin A and cos A, we change tan A to sin A/cos A and cot A to cos A/ sin A before further simplifying the expression.

(cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) + sinA + cosA

If (cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) + sinA + cosA = sin A + cos A

then,
(cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) = 0
which is also not correct

Your orginal question is to prove
cosA / (1 - tanA) + sinA / (1 - cotA) = Sin A + cosA

Let's focus on cosA / (1 - tanA) first. After changing tan A to sin A/cos A, what do you get ? (Hint: you need to simply the denominator first)

1-tan A
= 1- (sin A / cos A)
= (cos A - sin A) / cos A

therefore,
cos A / (1- tan A)
= cos A / [(cos A - sin A) / cos A]
= cos2A / (cos A - sin A)