# Trig Identities that I can't get a grip on

1. May 31, 2009

### Fractal314

[tan(pi/4+x)-tan(pi/4-x)]/[tan(pi/4+x)+tan(pi/4-x)]=2sinxcosx

I tried to prove this trig identity but I an really stuck. I think tan of pi/4 is '1', and if I do that then my numerator becomes zero, thus zero=2sinxcosx. But that can't be right, so I don't know what to do now.

LS= (1+tanx-1-tanx)/(1+tanx+1-tanx)

I get 0=2sinxcosx

Any thoughts? Also, I am wondering where this advanced formatting option is or how to do it.

2. May 31, 2009

### overt26

You will need to use the identities

tan(A+B)=(tanA+tanB)/(1-tanA*tanB)

tan(A-B)=(tanA-tanB)/(1+tanA*tanB)

3. May 31, 2009

### Fractal314

Thankyou Overt, I did not see it.

So would I be right in assigning 'y' as pi/4? and let 'x' be 'x'?

For tan(pi/4+x) I will instead get (tanx+tanpi/4)/(1-tanxtanpi/4)?

4. May 31, 2009

Correct!