Show that $\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$ Show that $$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$$
I have substituted the expansions for $\cos2x$ and $\sin2x$ and gotten, after simplification:
$$\frac{1-\sin x\cos x + 2\sin^2x}{1+\sin x\cos x-2\sin^2x}$$
I'm not sure how to carry on. I factored out the $\sin x$, but ended up with
$$\frac{1+\sin x}{1-\sin x}$$
I haven't been taught that as equal to $tan x$.

$$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \frac{1- (2\cos^2x-1)+2\sin x .\cos x}{1+ 2\cos^2x-1+2\sin x .\cos x}$$
$$ = \frac{2- 2\cos^2 x +2\sin x .\cos x}{ 2\cos^2 x +2\sin x .\cos x} = \frac{1- \cos^2 x +\sin x .\cos x}{ \cos^2 x + \sin x .\cos x}$$
$$ = \frac{\sin^2 x +\sin x .\cos x}{ \cos^2 x + \sin x .\cos x} = \frac{\sin x (\sin x + \cos x )}{ \cos x (\sin x + \cos x )} = \frac{\sin x }{ \cos x } = \tan x$$
