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Sophie [7]
3 years ago
5

PLEASE HELP ASAP 30 POINTS!!! WILL GIVE BRAINLIEST!!

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
5 0

1. The factoring step is

sin²<em>θ</em> - cos²<em>θ</em> sin²<em>θ</em> = sin²<em>θ</em> (1 - cos²<em>θ</em>)

Then the Pythagorean identity is invoked:

cos²<em>θ</em> + sin²<em>θ</em> = 1   →   1 - cos²<em>θ</em> = sin²<em>θ</em>

so that

sin²<em>θ</em> - cos²<em>θ</em> sin²<em>θ</em> = sin²<em>θ</em> sin²<em>θ</em> = sin⁴<em>θ</em>

(third option)

<em />

2. Recall that <em>a</em> ² - <em>b</em> ² = (<em>a</em> - <em>b</em>) (<em>a</em> + <em>b</em>). The numerator here is such a difference of squares:

csc²<em>x</em> - 1 = (csc<em>x</em> - 1) (csc<em>x</em> + 1)

Then

(csc²<em>x</em> - 1) / (1 + sin<em>x</em>) = ((csc<em>x</em> - 1) (csc<em>x</em> + 1)) / (1 + sin<em>x</em>)

Recall that csc<em>x</em> = 1/sin<em>x</em>, so rewrite this as

… = ((1/sin<em>x</em> - 1) (1/sin<em>x</em> + 1)) / (1 + sin<em>x</em>)

In the numerator, pull out a factor of 1/sin<em>x</em> from both terms:

… = (1/sin<em>x</em> (1 - sin<em>x</em>) × 1/sin<em>x</em> (1 + sin<em>x</em>)) / (1 + sin<em>x</em>)

… = ((1 - sin<em>x</em>) (1 + sin<em>x</em>)) / (sin²<em>x</em> (1 + sin<em>x</em>))

Cancel the common factor of 1 + sin<em>x</em> :

… = (1 - sin<em>x</em>) / sin²<em>x</em>

Expand the fraction and rewrite sin in terms of csc :

… = 1/sin²<em>x</em> - sin<em>x</em>/sin²<em>x</em>

… = 1/sin²<em>x</em> - 1/sin<em>x</em>

… = csc²<em>x</em> - csc<em>x</em>

Factor out csc<em>x</em> to get the second option,

… = csc<em>x</em> (csc<em>x</em> - 1)

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