Question

Where did I make the mistake? Suppose you prove the identity cot theta = csc theta/sec theta by working only on the write side of the equation. What would the final step look like? I only ask this because I had put; cot theta = 1/tan theta (since the reciprocal equals cot theta) but it was marked as inccorect. Would the final step just by cot theta = cot theta?

Answer 1

You know, it helps when you reduce *everything* to sines and cosines
I suggest you give that a try

Answer 2

Okay, hold on

Answer 3

I'm just a little confused because when I write the reciprocal identity for csc theta, it becomes a fraciton on top of another fraction. It that okay when solving for identities?

Answer 4

Of course, whenever you get a fraction on top of the other, you can express them as one *simpler* fraction. Do you know how?

Answer 5

I'm afraid not :(

Answer 6

Well, that's why I'm here ;)
When you get a fraction on top of another, say
(a/b) / (c/d)
you take the top fraction, and multiply it to the *reciprocal* of the bottom fraction
(a/b) * (d/c)
Can you take it from here and apply it to your problem?
:)

Answer 7

Right now, I have cot theta = (1/sin theta) / (1/cos theta), so I use the reciprocal of 1/cos theta and get theta? So it'd be cot theta = (1/sin theta) / cos theta? (:

Answer 8

I thought you were only working on the right side?

Answer 9

I ment * and get cos theta.

Answer 10

with csc theta / sec theta

Answer 11

Yes

Answer 12

Sorry Lol

Answer 13

ok, so csc = 1/sin, right? (I'll no longer put the theta)

Answer 14

what's sec?

Answer 15

Yes, and alrighty

Answer 16

You get it now? ;)

Answer 17

sec is 1/cos

Answer 18

A little more (:

Answer 19

that's right, so...
it boils down to
(1/sin) / (1/cos)
now apply what I showed you about complex fractions

Answer 20

So if I apply what you showed about the complex fractions, would it look like this? (1/sin) / (cos) ? I'm sorry if It's taking me a minute to grasp this.

Answer 21

ok, remember
(a/b) / (c/d)
= (a/b) * (d/c)
just substitute ;)

Answer 22

Ooohh, okay, didn't see the multiplication there (: Lol

Answer 23

So, can you do it?

Answer 24

Well, I have (1/sin) * cos but I'm not sure what follows...

Answer 25

Would it become tangent?

Answer 26

(1/sin) * (cos/1) = just multiply their numerators and denominators
and you should come to an epiphany :D

Answer 27

Ooooh, cotangnet? Because it'll be cos/sin, right?

Answer 28

And that proves the identity, right? (:

Answer 29

That's right
Good job :)

Answer 30

Thank you! I love how you made me think! Some people just give the answers and it doesn't help for future problems (: I appreciate your help!

Answer 31

Thanks, I learned my lesson :)
Uhh, it does help to try to reduce everything (if you can/want) to sines and cosines when proving identities

Answer 32

That's what my textbook said, but I didn't know what it really ment until now (: Thanks again!

Answer 33

No problem =D