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 one year 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
 one year 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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katherine.ok
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1376262907756:dw not covered in my book so i kinda know a little

katherine.ok
 one year 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)

katherine.ok
 one year ago
Best ResponseYou've already chosen the best response.0So the normal numerator of sin would be denomenator of csc , and etc..

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

tkhunny
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0so would the sec A= 15/9?

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

highschoolmom2010
 one year ago
Best ResponseYou've already chosen the best response.0sin (b)=9/15

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

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

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

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

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

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

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

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