(1-cos2x +sin2x) / (1+cos2x +sin2x) = tanx

\frac{1-cos2x +sin2x}{1+cos2x +sin2x} = tanx

\frac{1-(1-2sin^2x) +2sinxcosx}{1+(1-2sin^2x) +2sinxcosx} = tanx

\frac{2sin^2x +2sinxcosx}{2-2sin^2x +2sinxcosx} = tanx

\frac{2sinx(sinx +cosx)}{2(1-sin^2x) +2sinxcosx} = tanx

\frac{2sinx(sinx +cosx)}{2cos^2x+2sinxcosx} = tanx

\frac{2sinx(sinx +cosx)}{2cosx(cosx+sinx)} = tanx

\frac{2sinx(sinx +cosx)}{2cosx(sinx +cosx)} = tanx

Can you take that from here, it's already finished! :) Just remember, you have a single angle on the RHS, convert everything into single angles first. Then everything is simplifying.

I get everything up till where you went but I'm not to sure how to reduce to single form and simplify I'm reallly really confused with trig identities they are uber difficult!:(

But... but...

OK, I'll give you the answer. I'm really surprised you solved everything up to what I've done and yet, you couldn't do the rest, the actually simplest part.

the 2s are eliminated by division, as well as the (sinx+cosx)

\frac{2sinx(sinx+cosx)}{2cosx(sinx+cosx)}= tanx

\frac{sinx}{cosx}= tanx

Well, that's true, since \frac{sinx}{cosx}does equal to tanx

Wow I'm such an idiot I totally understand this stuff know! Just got a little brain fart yesterday

LOL, OK, that may happen. Just try to remember these if you have an assessment or a test or exams, k? :)

Now**