Question

Please help me with this "Proving Identities"

2/1+cos teta + 2/1-cos teta = 4 csc^2 teta

Answer 1

\[\frac{2}{1 + \cos(\theta)} + \frac{2}{1 - \cos(\theta)} = 4\csc^2(\theta)\]

Answer 2

^That's what it is, right?

Answer 3

@CHAsio_P Any guesses?

Answer 4

@Hero Yes.

Answer 5

@Hero yup
@ParthKohli none. how bout you?

Answer 6

Hint: Combine fractions on the left side.

Answer 7

Do you know how to get a common denominator?

Answer 8

@Hero thanks
@ParthKohli well, yeah. So it would be 1- cos \[\Theta r\]

Answer 9

You multiply through the denominators, and then you multiply the same to the numerator.\[\dfrac{2\color{#C00}{(1-\cos\theta)}}{\color{#C00}{(1 - \cos\theta)}(1 + \cos\theta)} + \dfrac{2\color{#C00}{(1 + \cos\theta)}}{(1 - \cos\theta)\color{#C00}{(1 + \cos\theta)}}\]

Answer 10

@ParthKohli how come that the denominator became like that? it is just so confusing. . .

Answer 11

Do you know that if you multiply both numerator and denominator by the same number (just not \(0\)), then you get an equivalent fraction.

Answer 12

@ParthKohli oh, thanks. then, i'll start doing the canceling?

Answer 13

No, you will add both fractions.

Answer 14

Before that, can you simplify \((1 - \cos\theta)(1 + \cos\theta)\)?

Answer 15

Honestly, no. Heheh. . .

Answer 16

Do you know that \((a - b)(a + b) = a^2 - b^2\)?

Answer 17

So, the denominator will be \[(1 - \cos ^2 \theta) + (1-\cos^2\theta)\]

Answer 18

?

Answer 19

Yes, you have\[\dfrac{2}{1 -\cos^2 \theta} + \dfrac{2}{1 - \cos^2 \theta}\]

Answer 20

When the denominators are equal, you can add the numerators. Consider this elementary example:\[\dfrac{2}{3} + \dfrac{3}{3} = \dfrac{2 + 3}{3} = \dfrac{5}{3}\]

Answer 21

So, it would be \[\frac{ 2 }{ 1-\cos ^2 }\]

Answer 22

?

Answer 23

Add the numerators...

Answer 24

@ParthKohli, see was asking if \[(1 - \cos ^2 \theta) + (1-\cos^2\theta)\] is the deominator. No it is not.

Answer 25

@Hero what do you mean??

Answer 26

@ParthKohli oh, \[\frac{ 4 }{ 1-\cos ^2\theta } ?\]

Answer 27

Yes! So we were checking if\[\dfrac{4}{1 - \cos^2 \theta} = 4\sec^2\theta\]Divide both sides by \(4\).

Answer 28

no, its not sec, its csc

Answer 29

I have a bad eyesight. Nvm. Can you divide both sides by \(4\)?

Answer 30

For the easiness of life,\[\dfrac{4}{1 - \cos^2 \theta} = 4\times\dfrac{1}{1 - \cos^2\theta}\]

Answer 31

I don't get it. How come that the \[4\csc^2\theta became 4 \times \frac{ 1 }{ 1-\cos^2\theta }?\]

Answer 32

My only question is, @CHAsio_P, what process you have to perform in order to find the denominator?

Answer 33

No, look you have:\[\frac{4}{1 -\cos^2\theta} = 4\csc ^2 \theta \]Look at the left hand side.

Answer 34

@ParthKohli Oh, sorry. Then, will I already chnage the \[4 \csc^2 \theta \to 4(\frac{ 1 }{ \sin^2\theta } ?\]

Answer 35

Yes :-)

Answer 36

And another thing: Do you know how you could simplify \(1 - \cos^2\theta\)?

Answer 37

There is a well known Pythagorean Identity associated with it.

Answer 38

Yup

Answer 39

@ParthKohli, I don't mean to interrupt, but when she asked:

"So is the denominator \((1 - \cos ^2 \theta) + (1-\cos^2\theta)\)", you agreed with her.

Answer 40

So, \[\frac{ 4 }{ \sin^2\theta } = 4 \frac{ 1 }{ \sin^2\theta } which is \frac{ 4 }{ \sin^2\theta } ?\]

Answer 41

Sorry, I said "Yup" to the line I wrote just after :-P

Answer 42

Yes @CHAsio_P!

Answer 43

@Hero its my mistake. Don't worry about it. :)) I already got it.

Answer 44

And how would you simplify\[\dfrac{4}{1-\cos^2\theta}\]

Answer 45

Oh my gosh! @ParthKohli thanks for your help!! :))))

Answer 46

What do you mean you already got it? I'm not convinced you have it.

Answer 47

You're welcome (I'm not really a great teacher!)

Answer 48

Well,its \[\frac{ 4 }{ \sin^2\theta } for 1 - \cos^2\theta is equals \to \sin^2\]

Answer 49

Right there.

Answer 50

@Hero it is just the denominator. I was just confuse, okay?

Answer 51

@ParthKohli thanks again!!!

Answer 52

No problem. :-D

Answer 53

So @CHASsio_P, can you explain in your own words, the process of finding the denominator?

Answer 54

Hmm. . . get the LCD. Heheh... I just dont know how to explain it by words, but i already got it. I swear. And I dont worry about the denominator, it is really my weakness. :))

Answer 55

Here's a simple question: How do you find the LCD?

Answer 56

Get the least common denominator or multiply the denominators .:))

Answer 57

So if you had

\[\frac{a}{b} + \frac{c}{d} = 5\] how do you find the LCD of that?

Answer 58

Oh, \[\frac{ ad }{ bd }+\frac{ ac }{ bd } ?Am I correct?\]

Answer 59

Yes, that is correct. Earlier you posted that for your original problem, the denominator should be \[(1 - \cos ^2 \theta) + (1-\cos^2\theta)\]. Can you correct your mistake here? What should that be instead?

Answer 60

Oh yeah. . . the sign, I am sorry okay. I was just confused. :))

Answer 61

Wait, actually, what you posted is not correct

Answer 62

\[\frac{ad}{bd} + \frac{ac}{bd}\] is not correct. And you made more than just a sign mistake for \[(1 - \cos ^2 \theta) + (1-\cos^2\theta)\]

Answer 63

First, for \[(1 - \cos ^2 \theta) + (1-\cos^2\theta)\], you should not be adding them together.

Answer 64

And then, yes you're right, there is a sign mistake. Instead of adding, multiply so what we should have for the denominator is \[(1 - \cos ^2 \theta) (1+\cos^2\theta)\] and of course that is equal to \[1 - \cos^2\theta\]

Answer 65

Okay. Thanks for explaining. :))

Answer 66

@Hero: Nice work :-)

Answer 67

Also for the other fraction, you want:

\[\frac{ad}{bd} + \frac{bc}{bd}\]

Answer 68

I think you might be able to figure things out from here, but for the future, hopefully, you'll try to understand how you might explain the process of finding an LCD or any math concept you've learned to others because only when you can explain it is when you have truly learned it.

Answer 69

I agree with you.

Answer 70

It's not necessarily enough to demonstrate computational ability. Being able to express the process verbally or written is the true measure of understanding.

Answer 71

But anyway, I'm done lecturing. Good luck to you :D

Answer 72

Yeah, thanks!! :))