anonymous
  • anonymous
Trigonometric Identities. Hello! I'd kindly ask for an explanation when solving for the following equation:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{ cotx+1 }{ cotx-1 } = \frac{ 1+tanx }{ 1-tanx}\]
anonymous
  • anonymous
well um equation um so you um yeah see ok hope that helped
anonymous
  • anonymous
I don't understand why the "1" and "-1" are replaced with\[\frac{ sinx }{ sinx }\]

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anonymous
  • anonymous
well bec you did the thing with that so it changed that thing *see*
anonymous
  • anonymous
May you please elaborate?
zepdrix
  • zepdrix
It's because you would like to get a common denominator when you switch your cotangents to sines and cosines :)
zepdrix
  • zepdrix
\[\large\rm \frac{\cot x+1}{\cot x-1}\quad=\frac{\frac{\cos x}{\sin x}+1}{\frac{\cos x}{\sin x}-1}\]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?
anonymous
  • anonymous
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?
zepdrix
  • zepdrix
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)
zepdrix
  • zepdrix
\[\large\rm \frac{\frac{a}{b}}{\frac{c}{d}-1}\quad\ne \frac{a}{b}\times \frac{d}{c}-1\]
anonymous
  • anonymous
Oh, that's right! So really, the 1's are substituted for a common denominator, correct?
zepdrix
  • zepdrix
Yes :) There is another approach though, which I think is much better, maybe you won't like it though hehe.
anonymous
  • anonymous
I'd love to know! I'm sure it won't be difficult to grasp C:
anonymous
  • anonymous
That is, if you don't mind.
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
And then the same for the bottom.
zepdrix
  • zepdrix
Oh I screwed up the signs on that last step of my long post. My bad.
anonymous
  • anonymous
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!
anonymous
  • anonymous
Is it because cosine is to be canceled?
anonymous
  • anonymous
(substituted for 1)
zepdrix
  • zepdrix
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
anonymous
  • anonymous
You're! Brilliant, I apologize for lack of communication thereof. I'm awfully exhausted, though, @zepdrix, thank you for your clarity, you're awesome!
zepdrix
  • zepdrix
Oh this one take all the energy out of ya? XD Haha ok np
anonymous
  • anonymous
Haha, slow process when learning Math c:

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