Maybe it should just be simplify????
tan x cos 2x + tan x + sin 2 x
\(tan x *cos 2x + tan x + sin 2 x\\ =\frac{sinx}{cosx} *(cos^2x-sin^2x)+ \frac{sinx}{cosx} + 2sinxcosx\\ =sinxcosx-\frac{sin^3x}{cosx}+ \frac{sinx}{cosx} + 2sinxcosx\\ =3sinxcosx-\frac{sin^3x}{cosx}+ \frac{sinx}{cosx} \\ =3sinxcosx- \frac{sinx}{cosx} (sin^2x-1)\\ =3sinxcosx- \frac{sinx}{cosx} (-cos^2x)\\ =3sinxcosx+ (sinxcosx)\\ =4sinxcosx\\ =2sin2x \)
Lets see if I could have done this more simply....
\( tan x cos 2x + tan x + sin 2 x\\ = tan x (cos^2x-sin^2x) + tan x + sin 2 x\\ = tan x (cos^2x-sin^2x+1) + sin 2 x\\ = tan x (cos^2x+cos^2x) + sin 2 x\\ =\frac{sinx}{cosx} (2cos^2x) + sin 2 x\\ =2sinxcosx + sin 2 x\\ =sin2x + sin 2 x\\ =2sin2x \\\)
So it can be seen that
tan x cos 2x + tan x = sin 2 x
which I guess was meant to be the question in the first place. (as hectictar and CPhill have already noted. Thanks guys. :)
tan x cos 2x + tan x + sin 2 x ......simplified ???
[sinx/ cosx] [ cos^2x - sin^2x] + sinx/cosx + 2sinx cosx
sinxcosx - sin^3x/ cosx + sinx/cosx + 2sinxcosx
3sinxcosx - [sinx ( sin^2x - 1)] / cosx
3sinxcosx - [ -sinx (1 - sin^2x) ]/ cosx
3sinxcosx + [sinx cos^2x]/cosx
3sinxcosx + sinxcosx =
4sinxcosx or 2sin(2x)