Trig Identity

**DaCoo911** · January 21st 2009, 03:29 PM

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

**o_O** · January 21st 2009, 03:34 PM

This looks awfully familiar to this identity: $\tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$

Let $a = x - y$ and $b = y$ and the conclusion follows.

**mollymcf2009** · January 21st 2009, 03:38 PM

Do you just need to simplify this using trig identities? Not sure what your question is.

**DaCoo911** · January 21st 2009, 03:54 PM

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.

**mr fantastic** · January 21st 2009, 09:37 PM

Originally Posted by DaCoo911

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.

Post #2 tells you exactly how to do it. If you're still stuck please say where.

**skeeter** · January 22nd 2009, 02:09 PM

hint ...

$\frac{\tan{a} + \tan{b}}{1 - \tan{a}\tan{b}} = \tan(a+b)$

**Jhevon** · January 22nd 2009, 02:13 PM

Originally Posted by DaCoo911

I have this identity in which i have to prove. Its supposed to be done by means of expanding the left side using sum and difference formulas and then simplifying down to tan x. Just can;t seem to figure it out. Thanks

recall, $\tan (x - y) = \frac {\tan x - \tan y}{1 + \tan x \tan y}$

apply this to the left hand side, then multiply the resulting expression by $\frac {1 + \tan x \tan y}{1 + \tan x \tan y}$ . can you take it from there?

**DaCoo911** · January 22nd 2009, 02:16 PM

yes that makes sense! thanks both of you for your help!

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