Question

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

Answer 1

Are you given or have you learned that csc= reciprocal of which , and sec of which , and cot is reciprocal of which?

Answer 2

not covered in my book so i kinda know a little

Answer 3

csc= sin inverse. cot= tan inverse, sec= cos inverse. (Sorry for the informal uses of words)

Answer 4

So the normal numerator of sin would be denomenator of csc , and etc..

Answer 5

so its like flipped

Answer 6

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.

\(\sin(B) = \dfrac{m(Opposite\;Side)}{m(Hypotenuse)} = \dfrac{1}{\csc(B)}\)

\(\cos(B) = \dfrac{m(Adjacent\;Side)}{m(Hypotenuse)} = \dfrac{1}{\sec(B)}\)

\(\tan(B) = \dfrac{m(Opposite\;Side)}{m(Adjacent\;Side)} = \dfrac{1}{\cot(B)}\)

Answer 7

SecA = 1/cosA

Find cosA first, this will let you find sec A.

CosA = adjacent/hypotenuse.

Answer 8

cos A= 9/15

Answer 9

Yes!!!

Answer 10

so would the sec A= 15/9?

Answer 11

Correct!

Now try cscB = 1/sinB

Sin = opp/hyp.

Answer 12

sin (b)=9/15

Answer 13

csc=15/9

Answer 14

Correct. Now try cotB. cotB = 1/tanB.

Answer 15

\[\tan (B)=\frac{ 9 }{ 12 }\]