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.

$\begin{align}\tan^2 x-\sin^2 x&=\sin^2 x(\sec^2 x-1) \\&=\sin^2 x\tan^2x\space \tag1\end{align}$

$$\tan 2x=\frac{2 \tan x}{1-\tan ^2x}$$
$$\tan 2x-2\tan x=\tan 2x\tan^2 x\space \tag2$$

From (1) and (2),
$$\begin{align}\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}&=\frac{\sin^2 x\tan^2x}{\tan 2x\tan^2 x}\\&=\frac{\sin^2x}{\tan 2x}\end{align}$$
which is equal to the given answer. (put $\sin^2x=\frac{1}{2}(1-\cos 2x)$)
