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,
,
Answer:
yes Wyatt is very correct
Actually the answer is
-3x + 5y = 15
2x + 3y = -10
tis noteworthy that the segment contains endpoints of A and C and the point B is in between A and C cutting the segment in a 1:2 ratio,
![\bf \textit{internal division of a line segment using ratios} \\\\\\ A(-9,-7)\qquad C(x,y)\qquad \qquad \stackrel{\textit{ratio from A to C}}{1:2} \\\\\\ \cfrac{A\underline{B}}{\underline{B} C} = \cfrac{1}{2}\implies \cfrac{A}{C}=\cfrac{1}{2}\implies 2A=1C\implies 2(-9,-7)=1(x,y)\\\\[-0.35em] ~\dotfill\\\\ B=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Binternal%20division%20of%20a%20line%20segment%20using%20ratios%7D%20%5C%5C%5C%5C%5C%5C%20A%28-9%2C-7%29%5Cqquad%20C%28x%2Cy%29%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bratio%20from%20A%20to%20C%7D%7D%7B1%3A2%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7BA%5Cunderline%7BB%7D%7D%7B%5Cunderline%7BB%7D%20C%7D%20%3D%20%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20%5Ccfrac%7BA%7D%7BC%7D%3D%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%202A%3D1C%5Cimplies%202%28-9%2C-7%29%3D1%28x%2Cy%29%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20B%3D%5Cleft%28%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22x%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cquad%20%2C%5Cquad%20%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22y%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cright%29%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf B=\left(\cfrac{(2\cdot -9)+(1\cdot x)}{1+2}\quad ,\quad \cfrac{(2\cdot -7)+(1\cdot y)}{1+2}\right)~~=~~(-4,-6) \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(2\cdot -9)+(1\cdot x)}{1+2}=-4\implies \cfrac{-18+x}{3}=-4 \\\\\\ -18+x=-12\implies \boxed{x=6} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(2\cdot -7)+(1\cdot y)}{1+2}=-6\implies \cfrac{-14+y}{3}=-6 \\\\\\ -14+y=-18\implies \boxed{y=-4}](https://tex.z-dn.net/?f=%5Cbf%20B%3D%5Cleft%28%5Ccfrac%7B%282%5Ccdot%20-9%29%2B%281%5Ccdot%20x%29%7D%7B1%2B2%7D%5Cquad%20%2C%5Cquad%20%5Ccfrac%7B%282%5Ccdot%20-7%29%2B%281%5Ccdot%20y%29%7D%7B1%2B2%7D%5Cright%29~~%3D~~%28-4%2C-6%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B%282%5Ccdot%20-9%29%2B%281%5Ccdot%20x%29%7D%7B1%2B2%7D%3D-4%5Cimplies%20%5Ccfrac%7B-18%2Bx%7D%7B3%7D%3D-4%20%5C%5C%5C%5C%5C%5C%20-18%2Bx%3D-12%5Cimplies%20%5Cboxed%7Bx%3D6%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B%282%5Ccdot%20-7%29%2B%281%5Ccdot%20y%29%7D%7B1%2B2%7D%3D-6%5Cimplies%20%5Ccfrac%7B-14%2By%7D%7B3%7D%3D-6%20%5C%5C%5C%5C%5C%5C%20-14%2By%3D-18%5Cimplies%20%5Cboxed%7By%3D-4%7D)
Answer:
Step-by-step explanation:
Translation of a point (h, k) by 'a' units to the right and 'b' units upwards is defined by,
(h, k) → (h + a, k +b)
Coordinates of A → (-4, -2)
Coordinates of B → (1, -1)
Coordinates of C → (0, -5)
If these points are shifted 4 units right and 3 units up,
By applying rules of the translation,
Coordinates of image point A' → (-4 + 4, -2 + 3)
→ (0, 1)
Coordinates of B' → (1 + 4, -1 + 3)
→ (5, 2)
Coordinates of C' → (0 + 4, -5 + 3)
→ (4, -2)
Now plot these points on the graph.