highschoolmom2010
  • highschoolmom2010
The sine, cosine, and tangent ratios each have a reciprocal ratio. The reciprocal ratios are cosecant (csc), secant (sec), and cotangent (cot). use triangle ABC to write the definitions sec A csc B cot B
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
Are you given or have you learned that csc= reciprocal of which , and sec of which , and cot is reciprocal of which?
highschoolmom2010
  • highschoolmom2010
|dw:1376262907756:dw| not covered in my book so i kinda know a little
anonymous
  • anonymous
csc= sin inverse. cot= tan inverse, sec= cos inverse. (Sorry for the informal uses of words)

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anonymous
  • anonymous
So the normal numerator of sin would be denomenator of csc , and etc..
highschoolmom2010
  • highschoolmom2010
so its like flipped
tkhunny
  • tkhunny
For the sake of clarity, let's not use "inverse" in this manner. This is a reciprocal. I can see why you used "inverse", but a few lessons down the road, you WILL regret it if you call it that. Pick an Angle!! Let's call it "B". It your drawing, it's one of the acute angles. After that, the definitions require memorization. You will be required to know which side is adjacent to the given acute angle and which side is opposite the given acute angle. \(\sin(B) = \dfrac{m(Opposite\;Side)}{m(Hypotenuse)} = \dfrac{1}{\csc(B)}\) \(\cos(B) = \dfrac{m(Adjacent\;Side)}{m(Hypotenuse)} = \dfrac{1}{\sec(B)}\) \(\tan(B) = \dfrac{m(Opposite\;Side)}{m(Adjacent\;Side)} = \dfrac{1}{\cot(B)}\)
agent0smith
  • agent0smith
SecA = 1/cosA Find cosA first, this will let you find sec A. CosA = adjacent/hypotenuse. |dw:1376263854146:dw|
highschoolmom2010
  • highschoolmom2010
cos A= 9/15
tkhunny
  • tkhunny
Yes!!!
highschoolmom2010
  • highschoolmom2010
so would the sec A= 15/9?
agent0smith
  • agent0smith
Correct! Now try cscB = 1/sinB Sin = opp/hyp. |dw:1376264245185:dw|
highschoolmom2010
  • highschoolmom2010
sin (b)=9/15
highschoolmom2010
  • highschoolmom2010
csc=15/9
agent0smith
  • agent0smith
Correct. Now try cotB. cotB = 1/tanB.
highschoolmom2010
  • highschoolmom2010
|dw:1376265085663:dw| \[\tan (B)=\frac{ 9 }{ 12 }\]
highschoolmom2010
  • highschoolmom2010
cot B= 12/9
highschoolmom2010
  • highschoolmom2010
@agent0smith do i have to simplify ^^
agent0smith
  • agent0smith
Yes, I would.
highschoolmom2010
  • highschoolmom2010
so the cot (b)= 4/3 csc (B)=5/3 sec (B)=5/3
agent0smith
  • agent0smith
Yep :)
highschoolmom2010
  • highschoolmom2010
cool i get it now :D ty
agent0smith
  • agent0smith
Oh except wasn't one of those sec A? secA was 5/3
highschoolmom2010
  • highschoolmom2010
@agent0smith i have A written down lol
highschoolmom2010
  • highschoolmom2010
ty @agent0smith @tkhunny & @katherine.ok for the help :D
agent0smith
  • agent0smith
Good job :)

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