Answer:
12x -y = -87
Step-by-step explanation:
You can start with the 2-point form of the equation of a line and manipulate it to give you the standard form.
y = (y2 -y1)/(x2 -x1)(x -x1) +y1
y = (-9 -3)/(-8 -(-7))(x -(-7)) +3
y = (-12/-1)(x +7) +3
y = 12x +84 +3
-12x +y = 87 . . . . . subtract the x-term
12x -y = -87 . . . . . . make the leading coefficient positive (per standard form)
Confused what the question is. Are you looking for the product or the zeroes?
If you are looking for the product, then:
Use foil to get: sec²(1) - sec²(-csc²) -1(1) -1(-csc²)
= sec² + sec²csc² - 1 + csc²
= sec²csc² + sec² + csc² - 1
= sec²csc² + 1 - 1 (NOTE: sec² + csc² = 1 is an identity)
= sec²csc²
Answer: sec²csc²
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If you are looking for the zeroes, then:
Using the zero product property, set each factor equal to zero and solve.
<u>First factor:</u>
sec²Θ - 1 = 0
sec²Θ = 1
secΘ = 1, -1
remember that secΘ is
= 1
= -1
cross multiply to get:
cosΘ = 1 cosΘ = -1
use the unit circle (or a calculator) to find that Θ = 0 and π
<u>Second factor:</u>
1 - csc²Θ = 0
1 = csc²Θ
1, -1 = cscΘ
remember that cscΘ is
= 1
= -1
cross multiply to get:
sinΘ = 1 sinΘ = -1
use the unit circle (or a calculator) to find that Θ = and
Answer: 0, π, ,