1. ## Trig 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!

2. 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.

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

4. 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.

5. 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.

6. hint ...

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

7. 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?

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