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highschoolmom2010
 3 years ago
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
highschoolmom2010
 3 years ago
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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Are you given or have you learned that csc= reciprocal of which , and sec of which , and cot is reciprocal of which?

highschoolmom2010
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1376262907756:dw not covered in my book so i kinda know a little

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0csc= sin inverse. cot= tan inverse, sec= cos inverse. (Sorry for the informal uses of words)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So the normal numerator of sin would be denomenator of csc , and etc..

highschoolmom2010
 3 years ago
Best ResponseYou've already chosen the best response.0so its like flipped

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.1For 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
 3 years ago
Best ResponseYou've already chosen the best response.1SecA = 1/cosA Find cosA first, this will let you find sec A. CosA = adjacent/hypotenuse. dw:1376263854146:dw

highschoolmom2010
 3 years ago
Best ResponseYou've already chosen the best response.0so would the sec A= 15/9?

agent0smith
 3 years ago
Best ResponseYou've already chosen the best response.1Correct! Now try cscB = 1/sinB Sin = opp/hyp. dw:1376264245185:dw

agent0smith
 3 years ago
Best ResponseYou've already chosen the best response.1Correct. Now try cotB. cotB = 1/tanB.

highschoolmom2010
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1376265085663:dw \[\tan (B)=\frac{ 9 }{ 12 }\]

highschoolmom2010
 3 years ago
Best ResponseYou've already chosen the best response.0@agent0smith do i have to simplify ^^

highschoolmom2010
 3 years ago
Best ResponseYou've already chosen the best response.0so the cot (b)= 4/3 csc (B)=5/3 sec (B)=5/3

highschoolmom2010
 3 years ago
Best ResponseYou've already chosen the best response.0cool i get it now :D ty

agent0smith
 3 years ago
Best ResponseYou've already chosen the best response.1Oh except wasn't one of those sec A? secA was 5/3

highschoolmom2010
 3 years ago
Best ResponseYou've already chosen the best response.0@agent0smith i have A written down lol

highschoolmom2010
 3 years ago
Best ResponseYou've already chosen the best response.0ty @agent0smith @tkhunny & @katherine.ok for the help :D
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