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 $tan x$.

\begin{align}
\frac{1-\cos 2x + \sin 2x}{1+\cos 2x + \sin 2x} &=\frac{1-(1-2\sin^2 x) + 2\sin x \cos x}{1+(2\cos^2 x -1) + 2\sin x \cos x} \\
&=\frac{2\sin^2x +2\sin x \cos x}{2\cos^2x + 2\sin x \cos x}
\end{align}
Can you simplify from here?
