1. The factoring step is
sin²θ - cos²θ sin²θ = sin²θ (1 - cos²θ)
Then the Pythagorean identity is invoked:
cos²θ + sin²θ = 1 → 1 - cos²θ = sin²θ
so that
sin²θ - cos²θ sin²θ = sin²θ sin²θ = sin⁴θ
(third option)
2. Recall that a ² - b ² = (a - b) (a + b). The numerator here is such a difference of squares:
csc²x - 1 = (cscx - 1) (cscx + 1)
Then
(csc²x - 1) / (1 + sinx) = ((cscx - 1) (cscx + 1)) / (1 + sinx)
Recall that cscx = 1/sinx, so rewrite this as
… = ((1/sinx - 1) (1/sinx + 1)) / (1 + sinx)
In the numerator, pull out a factor of 1/sinx from both terms:
… = (1/sinx (1 - sinx) × 1/sinx (1 + sinx)) / (1 + sinx)
… = ((1 - sinx) (1 + sinx)) / (sin²x (1 + sinx))
Cancel the common factor of 1 + sinx :
… = (1 - sinx) / sin²x
Expand the fraction and rewrite sin in terms of csc :
… = 1/sin²x - sinx/sin²x
… = 1/sin²x - 1/sinx
… = csc²x - cscx
Factor out cscx to get the second option,
… = cscx (cscx - 1)