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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

Mathematics
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Are you given or have you learned that csc= reciprocal of which , and sec of which , and cot is reciprocal of which?
|dw:1376262907756:dw| not covered in my book so i kinda know a little
csc= sin inverse. cot= tan inverse, sec= cos inverse. (Sorry for the informal uses of words)

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Other answers:

So the normal numerator of sin would be denomenator of csc , and etc..
so its like flipped
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. \(\sin(B) = \dfrac{m(Opposite\;Side)}{m(Hypotenuse)} = \dfrac{1}{\csc(B)}\) \(\cos(B) = \dfrac{m(Adjacent\;Side)}{m(Hypotenuse)} = \dfrac{1}{\sec(B)}\) \(\tan(B) = \dfrac{m(Opposite\;Side)}{m(Adjacent\;Side)} = \dfrac{1}{\cot(B)}\)
SecA = 1/cosA Find cosA first, this will let you find sec A. CosA = adjacent/hypotenuse. |dw:1376263854146:dw|
cos A= 9/15
Yes!!!
so would the sec A= 15/9?
Correct! Now try cscB = 1/sinB Sin = opp/hyp. |dw:1376264245185:dw|
sin (b)=9/15
csc=15/9
Correct. Now try cotB. cotB = 1/tanB.
|dw:1376265085663:dw| \[\tan (B)=\frac{ 9 }{ 12 }\]
cot B= 12/9
@agent0smith do i have to simplify ^^
Yes, I would.
so the cot (b)= 4/3 csc (B)=5/3 sec (B)=5/3
Yep :)
cool i get it now :D ty
Oh except wasn't one of those sec A? secA was 5/3
@agent0smith i have A written down lol
ty @agent0smith @tkhunny & @katherine.ok for the help :D
Good job :)

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