prove 1-cosx/1+cosx=(cosx-cscx)^2?

1-cosx/1+cosx=(cosx-cscx)^2
LHS=1-cosx/1+cosx=
=(1−cosx)(1−cosx)/{(1+cosx)(1−cosx)}
=(1−cosx)²/(1−cos²x)
=(1−cosx)²/sin²x
RHS =(cosx-cscx)²=(cosx-1/sinx)²
=(sinxcosx-1)²/sin²x
Hence LHS ≠RHS
Given identity is false

are you sure about the question, it isn't right when i substitute numbers into x

false identity.

Let x=pi/3
LHS = (1/2) / (3/2) =1/3

RHS = (1/2 - 2/sqrt(3))^2 =0.4286

Making assumptions about where parentheses should be,
(1 - cosx)/(1 + cosx) =
(1 - cosx)/(1 - cosx) * (1 - cosx)/(1 + cosx) =
(1 - 2cosx + cos^2x)/(1 - cos^2x) =
(1 - 2cosx + cos^2x)/sin^2x =

(cosx - cscx)^2 =
(cosx - 1/sinx)^2 =
cos^2x - 2cosx/sinx + 1/sin^2x =
cos^2x sin^2x/sin^2x - 2cosx sinx/sin^2x + 1/sin^2x =
(1 - 2cosx sinx + cos^2x sin^2x)/sin^2x

They're not equal.