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 $tan x$.
 A: Hint:
Use $$\cos2x=2\cos^2x-1=1-2\sin^2x\text{ and }\sin2x=2\sin x\cos x$$
Or use Weierstrass substitution
Or $$\dfrac{1+\sin2x}{\cos2x}=\dfrac{(\cos x+\sin x)^2}{\cos^2x-\sin^2x}=\dfrac{\cos x+\sin x}{\cos x-\sin x}\text { assuming }\cos x+\sin x\ne0$$
Apply Componendo et Dividendo
A: \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?
A: 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
$$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x}=\frac{1-\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}{1+\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}=\frac{1+t^2-1+t^2+2t}{1+t^2+1-t^2+2t}=\frac{2t+2t^2}{2+2t}=t$$
A: $$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \frac{1- (2\cos^2x-1)+2\sin x .\cos x}{1+ 2\cos^2x-1+2\sin x .\cos x}$$
$$ = \frac{2- 2\cos^2 x +2\sin x .\cos x}{ 2\cos^2 x +2\sin x .\cos x} = \frac{1- \cos^2 x +\sin x .\cos x}{ \cos^2 x + \sin x .\cos x}$$
$$ = \frac{\sin^2 x +\sin x .\cos x}{ \cos^2 x + \sin x .\cos x} = \frac{\sin x (\sin x + \cos x )}{ \cos x (\sin x + \cos x )} = \frac{\sin x }{ \cos x } = \tan x$$
