PLEASE HELP ASAP 30 POINTS 1. In the first step of proof, the left-handed side of the following identity is factored. Which fundamental identity is used in the second step of the proof?
2. Simplify the expression. csc^2x-1/1+sinx

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)

so, converting firstly the mixed fractions to improper fractions and then "adding".