# Thread: Need help with proving an identity!

1. ## Need help with proving an identity!

I need to verify this identity:

(tan x/(1 + cos x)) + (sin x/(1 - cos x)) = cot x + (sec s)(csc x)

I've tried it twice now, and on the left side I keep getting:
sin^2 x /(cot x + (sec x)(csc x)

What I'm doing is multiplying the denominator and the numerator of the first fraction by (1 - cos x) to get sin x^2 on the bottom, and doing the same with the second fraction by multiplying by (1 + cos x). Simplifying that give me the right half of the identity under sin ^2 X

It's supposed to be an identity but I guess I'm screwing it up... can someone try it out and post their work so I can see what I'm doing wrong?

I'd very much appreciate it.

2. Hi

The idea is the right one :
$\frac{\tan x}{1+\cos x}+\frac{\sin x}{1-\cos x}=\frac{\frac{\sin x}{\cos x}(1-\cos x)+(1+\cos x)\sin x}{\sin^2x}=\frac{\frac{1-\cos x}{\cos x}+1+\cos x}{\sin x}=\ldots$

3. Thanks... but I dont quite understand what you are doing between the second and third step, and I don't know what to do after that....

4. Hello,

Divided by sin(x)

5. Originally Posted by flyingsquirrel
Hi

The idea is the good one :
$\frac{\tan x}{1+\cos x}+\frac{\sin x}{1-\cos x}=\frac{\frac{\sin x}{\cos x}(1-\cos x)+(1+\cos x)(\sin x)}{\sin^2x}=\frac{\frac{1-\cos x}{\cos x}+1+\cos x}{\sin x}=\ldots$
THe good idea?

Who would have guessed you were French? Bon idee!

6. Originally Posted by Mathstud28
THe good idea?

Who would have guessed you were French? Bon idee!
Are you saying that French people can't guess correctly ?

7. Originally Posted by Moo
Are you saying that French people can't guess correctly ?
Mais non!

8. Originally Posted by Kiba
Thanks... but I dont quite understand what you are doing between the second and third step, and I don't know what to do after that....
$\frac{1-\cos x}{\cos x}=\frac{1}{\cos x}-1$ then simplify in the last expression of my first post and you got the result.
Originally Posted by Mathstud28
THe good idea?
Is there another one ?
Bonne idee!

9. Originally Posted by flyingsquirrel
$\frac{1-\cos x}{\cos x}=\frac{1}{\cos x}-1$ then simplify in the last expression of my first post and you got the result.

Is there another one ?
Crap...and $\text{bonNE idee}\Rightarrow{right} \text{idea}$
Your sentence would be "This idea is the correct/right one"
Dont worry...today I said ....C'est possible ce(instead of que) je n'ais pas raison"

10. I still don't get it... and I don't know how to simplify that anymore...

11. Originally Posted by Mathstud28
Crap...and $\text{bonNE idee}\Rightarrow{right} \text{idea}$
Your sentence would be "This idea is the correct/right one"
Dont worry...today I said ....C'est possible ce(instead of que) je n'aie pas raison"
As for the problem:
$\frac{\frac{\sin x}{\cos x}(1-\cos x)+(1+\cos x)\sin x}{\sin^2x}$

Factor out sin x from the numerator and cancel with the denominator:
$= \frac{\frac{1}{\cos x} - 1 + 1 + \cos x}{\sin x}$

Simplify:
$= \frac{\sec x + \cos x}{\sin x} = \frac{\sec x}{\sin x} + \frac{\cos x}{\sin x}$

I'm sure you can see what happen sfrom here.

12. Originally Posted by o_O
As for the problem:
$\frac{\frac{\sin x}{\cos x}(1-\cos x)+(1+\cos x)\sin x}{\sin^2x}$

Factor out sin x from the numerator and cancel with the denominator:
$= \frac{\frac{1}{\cos x} - 1 + 1 + \cos x}{\sin x}$

Simplify:
$= \frac{\sec x + \cos x}{\sin x} = \frac{\sec x}{\sin x} + \frac{\cos x}{\sin x}$

I'm sure you can see what happen sfrom here.
Mon dieu...quest-ce que tu fait? Arrette! Je sais. Je ne suis pas bon a Francais...fiche-moi!

13. Originally Posted by Mathstud28
Mon dieu...qu'est-ce que tu fais? Arrête! Je sais. Je ne parle pas bien la français...fiche-moi le camp (?)!

Moo can fix anything else

14. He has quite a rude language in French :/ Not only here I mean..
"le français"
or "fiche-moi la paix"

It's perfect