Question

Trigonometric Identities.

Hello! I'd kindly ask for an explanation when solving for the following equation:

Answer 1

\[\frac{ cotx+1 }{ cotx-1 } = \frac{ 1+tanx }{ 1-tanx}\]

Answer 2

well um equation um so you um yeah see ok hope that helped

Answer 3

I don't understand why the "1" and "-1" are replaced with\[\frac{ sinx }{ sinx }\]

Answer 4

well bec you did the thing with that so it changed that thing *see*

Answer 5

May you please elaborate?

Answer 6

It's because you would like to get a common denominator when you switch your cotangents to sines and cosines :)

Answer 7

\[\large\rm \frac{\cot x+1}{\cot x-1}\quad=\frac{\frac{\cos x}{\sin x}+1}{\frac{\cos x}{\sin x}-1}\]If you look at the denominator of the main fraction,
see how it's difficult to deal with these two terms because one is a fraction and the other is just 1?

Answer 8

Oh, I was in the equation panel. Didn't see that you've already drawn the scenario. Well, wouldn't the 1's just cancel when I flip the denominator?

Answer 9

No no no, you can't flip until you have a single fraction in the top and bottom.
(If that's what you were implying)

Answer 10

\[\large\rm \frac{\frac{a}{b}}{\frac{c}{d}-1}\quad\ne \frac{a}{b}\times \frac{d}{c}-1\]

Answer 11

Oh, that's right! So really, the 1's are substituted for a common denominator, correct?

Answer 12

Yes :)
There is another approach though, which I think is much better,
maybe you won't like it though hehe.

Answer 13

I'd love to know! I'm sure it won't be difficult to grasp C:

Answer 14

That is, if you don't mind.

Answer 15

Instead what you `could do` is multiply the top and bottom by sin x.\[\large\rm \frac{\cot x+1}{\cot x-1}\quad=\frac{\frac{\cos x}{\sin x}+1}{\frac{\cos x}{\sin x}-1}\left(\frac{\sin x}{\sin x}\right)\]The reason I chose sin x is because that's the only thing showing up in the denominator of these fractions,
So I'm multiplying top and bottom by sin x to get rid of the denominators.\[\large\rm =\frac{\cos x+\sin x}{\cos x-\sin x}\]Oh oh oh,
but you're trying to turn this into tangents.
Ahh my bad.
That would require another step then, dividing top and bottom by cos x,\[\large\rm =\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}\]And I suppose it's `not` easier at that point :)

Maybe we just stick to what you were doing hehe.

Answer 16

In case there was any confusion on what I was doing,\[\large\rm \left(\frac{\cos x}{\sin x}+1\right)\sin x\quad=\frac{\cos x}{\sin x}\sin x+\sin x\]I was giving the top sin x to each term in the numerator, distributing like this.

Answer 17

And then the same for the bottom.

Answer 18

Oh I screwed up the signs on that last step of my long post.
My bad.

Answer 19

Aha! :) Also, when the denominator is flipped, how come there's a \[\frac{ 1 }{ \cos}\]?

For example, \[\frac{ \frac{ 1 }{ cosx } }{ \frac{ 1 }{ cosx } } * \frac{ \cos+\sin }{ \cos-\sin }\]

Also, thank you for the help thus far, I really do appreciate it!

Answer 20

Is it because cosine is to be canceled?

Answer 21

(substituted for 1)

Answer 22

I'm not sure what you mean by `flipped`.
Keep in mind that your `Keep Change Flip` (or however you learned this trick)
is just something that you can `choose` to do.

It's not required.
So for your example:\[\large\rm \frac{\frac{1}{\cos x}}{\frac{1}{\cos x } } \cdot \frac{\cos x+\sin x}{\cos x-\sin x}=\frac{\frac{1}{cos x}(cos x+sin x)}{\frac{1}{cos x}(cos x-sin x)}\]And then yes, distributing the 1/cos x to each term gives you a cosx/cosx as one of your terms, which simplifies to 1.

Maybe I'm not understanding what you're asking though :3

Answer 23

You're! Brilliant, I apologize for lack of communication thereof. I'm awfully exhausted, though, @zepdrix, thank you for your clarity, you're awesome!

Answer 24

Oh this one take all the energy out of ya? XD
Haha ok np

Answer 25

Haha, slow process when learning Math c: