Other approaches to simplify $\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}$ I want to simplify the trigonometric expression $\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}$.
My approach,
Here I used the abbreviation $s,c,t$ for $\sin x$ and $\cos x$ and $\tan x$ respectively,
Numerator is, $$\frac{s^2}{c^2}-s^2=\frac{s^2-s^2c^2}{c^2}=\frac{s^4}{c^2}=s^2t^2.$$
And denominator is,
$$\frac{2t}{1-t^2}-2t=\frac{2t^3}{1-t^2}.$$
So $$\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}=s^2t^2\times\frac{1-t^2}{2t^3}=\sin^2x\times \frac1{\tan 2x}=\frac{1-\cos 2x}{2\tan 2x}.$$
I'm looking for alternative approaches to simplify the expression.

My method uses Euler's formula and Computer Algebra Systems.
Let $\,X := e^{i\,x}.\,$ Then
$$ s := \sin(x) = \frac{X-X^{-1}}{2i}, \;\;
   t := \tan(x) = \frac{X-X^{-1}}{i(X+X^{-1})}, \\
   c := \cos(x) =  \frac{X+X^{-1}}{2}, \;\;
   t_2 := \tan(2x) = \frac{X^2-X^{-2}}{i(X^2+X^{-2})}. $$
Simplify a little to get
$$ s = \frac{1-X^2}{2iX}, \;\;
   c = \frac{1+X^2}{2X}, \;\;
   t =  \frac{1-X^2}{i(1+X^2)}, \;\;
   t_2 = \frac{1-X^4}{i(1+X^4)}. $$
Now get with Computer Algebra System factoring
$$ f := \frac{t^2 - s^2}{t_2-2t} =
   \frac{(1 - X^2)(1 + X^4)}{4i(1 + X^2)X^2}. $$
Rearrange the factors of the expression as
$$ f = \frac{(1-X^2)}{i(1+X^2)}\frac{(1+X^4)}{4X^2}
 = \frac{t\,c_2}2 = \frac{\tan(x)\cos(2x)}2 . $$
Alternatively rewrite it as
$$ f = \frac{(1-X^2)^2}{4i^2X^2}
  \frac{i(1+X^4)}{(1-X^2)(1+X^2)} = \frac{s^2}{t_2} 
= \frac{\sin(x)^2}{\tan(2x)}.$$
Yet another alternative is
$$ \frac18\frac{(1-X^2)(1+X^2)(1+X^4)}{2iX^4}
  \frac{4X^2}{(1+X^2)^2} = \frac{s_4}{8c^2} =
  \frac{\sin(4x)}{8\cos(x)^2}.
$$
Which one of these alternatives is simpler than the
others is not clear unless a precise definition of
simplicity is given in this context.
