Question

Prove:

Answer 1

Prove what?

Answer 2

(Cotx+Tanx) / Secx = Cscx

Answer 3

Do you know the definition of tangent and cotangent? Like, how to write them in terms of sin and cos?

Answer 4

Yes

Answer 5

Ok, do that first!

Answer 6

i did but then i got stuck

Answer 7

Get unstuck. :D

Answer 8

im not sure how

Answer 9

Write what you have.

Answer 10

((Cosx/Sinx) + (Sinx/Cosx)) / (1/Cosx)

Answer 11

So to be able to add \(\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}\) we need a common denominator. Remember that you can multiply anything by 1 and anything over itself is equal to 1.
\[\frac{\cos x}{\sin x}(\frac{\cos x}{\cos x}) + \frac{\sin x}{\cos x}(\frac{\sin x}{\sin x})\]

Answer 12

Alright that makes sense but after i multiply what do i do?

Answer 13

Add and simplify.
\[\frac{\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}}{\frac{1}{\cos x}}\]\[\frac{\frac{1}{\sin x\cos x}}{\frac{1}{\cos x}}\]\[\frac{\cos x}{\sin x\cos x}\]\[\frac{1}{\sin x}\]
Which is of course the same as \(\csc x\).

Answer 14

Whoa alright that makes complete sense now thanks!

Answer 15

Alright i got another one for ya

Answer 16

Ok...

Answer 17

(CosxTanx + 2Cosx - Tanx - 2) / (Tanx + 2) = Cosx -1

Answer 18

Im not sure where to start ...

Answer 19

Write things in terms of more basic expressions and simplify, just like the last one.

Answer 20

So like ((1/Secx)(1/Cotx) + 2(1/Secx) - (1/Cotx) - 2) / ((1/Cotx) + 2)

Answer 21

Well, that's not more basic. I meant more along the lines of
\[\cos x\tan x = \cos x\frac{\sin x}{\cos x} = \sin x\]

Answer 22

Oh okay, what about 2Cosx - Tanx - 2 ?

Answer 23

This is the part where you do a bit instead of me just giving you the answer.

Answer 24

I know im thinking and trying to figure it out but i cant figure out what 2Cosx would simplify to. Would it be (2/1)(Cosx/1) ?

Answer 25

You have to think bigger. Look at the whole expression.
\[\frac{\sin x + 2\cos x + \tan x - 2}{\tan x + 2}\ = \cos x - 1\]
So how can we get rid of some of that? How about like this:
\[\frac{\sin x + 2\cos x + \tan x - 2}{\tan x + 2}\ = \cos x - \frac{\tan x + 2}{\tan x + 2}\]

Answer 26

Oops, that should be \(\sin x + 2\cos x - \tan x - 2\) in the numerator.

Answer 27

Okay but why are you messing with the RHS ?

Answer 28

Because that helps us simplify the expression and get rid of some of the stuff.

Answer 29

Alright but how does that help simplify it ? I dont understand

Answer 30

Because now we have something we can add to the left side to get rid of some stuff.
Add \(\frac{\tan x + 2}{\tan x + 2}\) to both sides and show me what you get.

Answer 31

((Sinx + 2Cosx) / 2( Tanx + 2)) = Cosx

Answer 32

Uh, what's \(\frac{1}{7} + \frac{1}{7}\)?

Answer 33

2/7

Answer 34

Right. Not \(\frac{2}{14}\). So check your work.

Answer 35

OH okay, ((Sinx + 2Cosx) / (Tanx +2)) = Cosx

Answer 36

Yes.
Now you see why I did that.

Answer 37

not really

Answer 38

It got rid of stuff! It made it simpler.

Answer 39

okay yeah

Answer 40

So keep going. What's the next step?

Answer 41

i have no idea

Answer 42

What about that expression is getting in the way of simplifying it further?

Answer 43

what?

Answer 44

Maybe it'd help to get rid of the denominator on the left side.

Answer 45

im confused

Answer 46

\[\frac{\sin x + 2\cos x}{\tan x + 2} = \cos x\]
Get rid of the denominator on the left side.

Answer 47

Would you multiply by the conjugate?

Answer 48

How does that even help?

Answer 49

idk

Answer 50

i have to go to sleep though i got class in the am

Answer 51

Ok. Bye.

Answer 52

You're welcome. :|