The given identity is,

`(tan(A))/(sec(A)-1) - (sin(A))/(1+cos(A)) = 2cot(A)`

I will start from LHS:

LHS =

`= (tan(A))/(sec(A)-1) - (sin(A))/(1+cos(A))`

I can rewrite sec(A) and tan(A) as,

`= ((sin(A))/(cos(A)))/((1/(cos(A)))-1) - (sin(A))/(1+cos(A))`

Simplifying,

`= (sin(A))/(1-cos(A)) - (sin(A))/(1+cos(A))`

Taking both terms into a common denominator,

`= (sin(A)(1+cos(A))-(sin(A)(1-cos(A))))/((1-cos(A))(1+cos(A)))`

=This gives,

`= (sin(A)+sin(A)cos(A)-sin(A)+sin(A)cos(A))/(1-cos^2(A))`

`= (2sin(A)cos(A))/(sin^2(A))`

(Because,` sin^2(A)+cos^2(A) =1` )

LHS =

`= (2cos(A))/(sin(A))`

`= 2cot(A)`

Therefore, LHS = RHS, and

`(tan(A))/(sec(A)-1) - (sin(A))/(1+cos(A)) = 2cot(A)`           Proved.