<s>[INST] 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 $tan x$.
 [/INST] By Tangent half-angle formulas with $t=\tan x$

*

*$\cos 2x=\frac{1-t^2}{1+t^2}$

*$\sin 2x=\frac{2t}{1+t^2}$
we have
$$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x}=\frac{1-\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}{1+\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}=\frac{1+t^2-1+t^2+2t}{1+t^2+1-t^2+2t}=\frac{2t+2t^2}{2+2t}=t$$
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