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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
 8 months ago
 8 months 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
 8 months ago
 8 months ago

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katherine.okBest ResponseYou've already chosen the best response.0
Are you given or have you learned that csc= reciprocal of which , and sec of which , and cot is reciprocal of which?
 8 months ago

highschoolmom2010Best ResponseYou've already chosen the best response.0
dw:1376262907756:dw not covered in my book so i kinda know a little
 8 months ago

katherine.okBest ResponseYou've already chosen the best response.0
csc= sin inverse. cot= tan inverse, sec= cos inverse. (Sorry for the informal uses of words)
 8 months ago

katherine.okBest ResponseYou've already chosen the best response.0
So the normal numerator of sin would be denomenator of csc , and etc..
 8 months ago

highschoolmom2010Best ResponseYou've already chosen the best response.0
so its like flipped
 8 months ago

tkhunnyBest ResponseYou've already chosen the best response.1
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. \(\sin(B) = \dfrac{m(Opposite\;Side)}{m(Hypotenuse)} = \dfrac{1}{\csc(B)}\) \(\cos(B) = \dfrac{m(Adjacent\;Side)}{m(Hypotenuse)} = \dfrac{1}{\sec(B)}\) \(\tan(B) = \dfrac{m(Opposite\;Side)}{m(Adjacent\;Side)} = \dfrac{1}{\cot(B)}\)
 8 months ago

agent0smithBest ResponseYou've already chosen the best response.1
SecA = 1/cosA Find cosA first, this will let you find sec A. CosA = adjacent/hypotenuse. dw:1376263854146:dw
 8 months ago

highschoolmom2010Best ResponseYou've already chosen the best response.0
so would the sec A= 15/9?
 8 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Correct! Now try cscB = 1/sinB Sin = opp/hyp. dw:1376264245185:dw
 8 months ago

highschoolmom2010Best ResponseYou've already chosen the best response.0
sin (b)=9/15
 8 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Correct. Now try cotB. cotB = 1/tanB.
 8 months ago

highschoolmom2010Best ResponseYou've already chosen the best response.0
dw:1376265085663:dw \[\tan (B)=\frac{ 9 }{ 12 }\]
 8 months ago

highschoolmom2010Best ResponseYou've already chosen the best response.0
@agent0smith do i have to simplify ^^
 8 months ago

highschoolmom2010Best ResponseYou've already chosen the best response.0
so the cot (b)= 4/3 csc (B)=5/3 sec (B)=5/3
 8 months ago

highschoolmom2010Best ResponseYou've already chosen the best response.0
cool i get it now :D ty
 8 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Oh except wasn't one of those sec A? secA was 5/3
 8 months ago

highschoolmom2010Best ResponseYou've already chosen the best response.0
@agent0smith i have A written down lol
 8 months ago

highschoolmom2010Best ResponseYou've already chosen the best response.0
ty @agent0smith @tkhunny & @katherine.ok for the help :D
 8 months ago
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