Trig Identity: Can you prove that tan(x) = sin(2x)/[1+cos(2x)]. Thanks?

tan(x) = sin(2x)/[1+cos(2x)]

sin(2x) = 2sinxcosx;

cos(2x) = cos^2(x) -sin^2(x)

tan(x) = 2sinxcosx/[1+cos^2(x) -sin^2(x)]

tan(x) = 2sinxcosx/[1+cos^2(x) -1 +cos^2(x)] <= sin^2(x) = 1 - cos^2(x)

tan(x) = 2sinxcosx/[2cos^2(x)]

tan(x) = sinx/[cos(x)]

tan(x) = tan(x);

RHS = (tan pi/4 - tan x)/(a million+ tan x tan pi/4) = (a million-tan x)/(a million+an x) = (cos x -sin x)/(sin x+ cos x) = ( cos x- sin x)(sin x + cos x)/(sin x+ cos x)^2 = (cos ^2 x -sin ^2 x)/(sin ^2 x+ cos^2 x + 2 sin x cos x) = cos 2x/(a million+ sin 2x)

tan (x) = sin (2x)/[1+cos (2x)]

forms of equation that may derived:

sin (2x) = 2sin x cos x

cos (2x) = 2cos² x - 1

tan x = [2sin x cos x] / [1+2cos² x -1]

tan x = [2sin x cos x] / [1-1+2cos² x]

tan x = [2sin x cos x] / [2cos² x] ..... cancel the 2 of

numerator and denominator gives you...

tan x = [sin x cos x] / [cos² x] ......... divide the numerator and denominator.

tan x = sin x/cos x

tan x = tan x

RHS=2sinxcosx/(1+2cos^2x-1)

=2sinxcosx/2cos^2x

=sinx/cosx=tanx=LHS

x = 6

sin(2x)/[1+cos(2x)= 2 sin(x) cos(x) / 1+1-2sin^2(x)

using sin(2x)=2 sin(x) cos(x)

and cos(2x)=1-2sin^2(x)

=2 sin(x) cos(x) / 2-2sin^2(x)

=2 sin(x) cos(x) / 2(1-sin^2(x))

=2 sin(x) cos(x) / 2 cos^2(x) = sin(x)/cos(x)

=tan(x)