HELLLLLP! lol i've tried for soo long now. I'm still getting used to these things...they're just like a PUZZLE! thanks!

http://i546.photobucket.com/albums/h...911/Help-1.jpg

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- Jan 21st 2009, 03:29 PMDaCoo911Trig Identity
HELLLLLP! lol i've tried for soo long now. I'm still getting used to these things...they're just like a PUZZLE! thanks!

http://i546.photobucket.com/albums/h...911/Help-1.jpg - Jan 21st 2009, 03:34 PMo_O
This looks awfully familiar to this identity: $\displaystyle \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$

Let $\displaystyle a = x - y$ and $\displaystyle b = y$ and the conclusion follows. - Jan 21st 2009, 03:38 PMmollymcf2009
Do you just need to simplify this using trig identities? Not sure what your question is.

- Jan 21st 2009, 03:54 PMDaCoo911
ya its just a matter of expanding the left side and simplyfiying down to tanx, i just can't find a way to do it.

- Jan 21st 2009, 09:37 PMmr fantastic
- Jan 22nd 2009, 02:09 PMskeeter
hint ...

$\displaystyle \frac{\tan{a} + \tan{b}}{1 - \tan{a}\tan{b}} = \tan(a+b)$ - Jan 22nd 2009, 02:13 PMJhevon
- Jan 22nd 2009, 02:16 PMDaCoo911
yes that makes sense! thanks both of you for your help!