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.

There are many ways to do it.
For axample, let $x=\tan^{-1}(t)$ and, after simplifications
$$\frac{\tan ^2(x)-\sin ^2(x)}{\tan (2 x)-2 \tan (x)}=\frac{1}{2} t \cos \left(2 \tan ^{-1}(t)\right)=\frac{t \left(1-t^2\right)}{2 \left(1+t^2\right)}=\frac 1 2 \times t\times \frac{ \left(1-t^2\right)}{ \left(1+t^2\right)}$$ that is to say
$$\frac 1 2 \times \tan(x) \times \cos(2x)$$
Another way is to use $\tan(2x)=\frac{2 \tan (x)}{1-\tan ^2(x)}$
