Show that function is strictly monotone increasing

let $f(x)=\dfrac{x-\sin(x)}{1-\cos(x)}$

EDIT: I am wrong in the formula  for the last version and I correct here,sorry for it.

$f'(x)=\dfrac{(x-\sin x)(-\sin x)}{(1-\cos x)^2}+\dfrac{1-\cos x}{1-\cos x}=\dfrac{\sin^2x-x\sin x+1-2\cos x+\cos^2 x}{(1-\cos x)^2}=\dfrac{2(1-\cos x)-x\sin x}{(1-\cos x)^2}=\dfrac{2\cdot 2\sin^2 \dfrac{x}{2}-2x\sin\dfrac{x}{2}\cos{x}{2}}{(1-\cos x)^2}=\dfrac{2\cdot sin \dfrac{x}{2}}{(1-\cos x)^2}\cdot (2\sin\dfrac{x}{2}-x\cos x \dfrac{x}{2}) $

$\dfrac{2\cdot \sin \dfrac{x}{2}}{(1-\cos x)^2}>0 $ when $0<x<2\pi$,

so we only check $(2\sin\dfrac{x}{2}-x\cos x \dfrac{x}{2}) $.

When $0< x < \pi, \tan\dfrac{x}{2}>\dfrac{x}{2},$ we have $ 2\sin\dfrac{x}{2}>x\cos x \dfrac{x}{2}$, ie. $2\sin\dfrac{x}{2}-x\cos x \dfrac{x}{2}>0$.

When $\pi \leq x < 2\pi$, $\cos\dfrac{x}{2} \leq 0, -x \cos\dfrac{x}{2} \ge 0,$ and $2\sin\dfrac{x}{2}>0$, ie. $2\sin\dfrac{x}{2}-x\cos x \dfrac{x}{2}>0$

so $f'(x)>0$ when $0<x<2\pi$, QED

As ᴊᴀsᴏɴ said, you can

Note that the denominator is actually the derivative of the numerator.

Which can be written as $f(x)=\frac{u}{u'}$ and we know that $\frac{d \ln(u)}{dx}=\frac{u'}{u}$ which represents the slope of the $\ln$ function on the interval mapped by $u(x)$ where $x\in]0,2\pi[$

Now think about it, the slope of the $ln$ function is strictly decreasing, so you can see your function as the inverse of this slope and thus strictly increasing.

The derivative is $$f'(x)=\frac{2(1-\cos(x))-x \sin(x)}{(1-\cos(x))^2}=\frac{4\sin^2 \frac{x}{2}-2x\cos \frac{x}{2} \sin \frac{x}{2}}{(1-\cos(x))^2}=\frac{2\sin \frac{x}{2}(2\sin \frac{x}{2}-x \cos \frac{x}{2})}{(1-\cos(x))^2}$$

All terms here are non-negative on $(0,2\pi)$, except possibly $2 \sin \frac{x}{2}-x \cos \frac{x}{2}$. We will prove that it is nonnegative as well. i.e. $$2 \sin \frac{x}{2} \geq x \cos \frac{x}{2}$$ For $x \in[\pi,2\pi)$ is it obvious, because the RHS is non-positive, white the LHS is non-negative.

For $x \in (0,\pi)$, divide by the cosine, to get $2 \tan \frac{x}{2}-x\geq0$, which follows from the identity $\tan t \geq t$ for $t \in(0,\frac{\pi}{2})$.

In summary, we proved $f' \geq 0$, as required.