Express this trigonometric identity in simplest form?

(csc^2x - 1)/(cscx - 1) =
(cscx + 1)(cscx - 1)/(cscx - 1) =
cscx + 1

(csc^2 (x) - 1)/(csc (x) - 1)
= csc (x) + 1

It's not an identity.

csc(x) + 1, provided that sin(x) =/= 1. Simply factor the numerator as a difference of squares, then cancel out.

[csc^2(x)-1]/[csc(x) - 1] = csc(x) + 1]*[ csc(x) - 1]/ [csc(x) - 1],
= csc(x) + 1,
or, = [1 + sin(x)]/sin(x) ...........(Ans.)

This is a wrongly worded question. It quotes an expression not an identity. An identity has an equals sign in it.
The question should read "simplify the expression...."

you have an " [ a² - b² ] / [ a - b ] "...use it

Rule
a^2-b^2=(a-b)(a+b)

(csc^2 (x) - 1)/(csc(x) - 1)
= ((cos (x) + 1)(cos (x) - 1))/(csc(x) - 1)
= csc(x) + 1

[ csc x - 1 ] [ csc x + 1 ]
-------------------------------- = csc x + 1
csc x - 1