Show that function is strictly monotone increasing I want to show that $$ f(x)=\dfrac{x-\sin(x)}{1-\cos(x)} $$ is strictly increasing in $(0,2 \pi) $. Unforunately, this is not that easy for me , as the derivative is not very manageable and trigonometric identities appear to be not very helpful, too.$$$$
Are there any further ways to do this? 
 A: 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
A: 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.
A: 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.
A: If you are unable to differentiate (though , I'll suggest differentiating . It would be easy to prove).
You can draw the curve for this function.

You can see that function increases monotonically in $[0,2\pi]$.(Aproaching to $\infty$ at $x=2\pi$)
Else. $$f'(x)=\dfrac{2(1-\cos x)-x\sin x}{(1-\cos x)^2}$$
Numerator(N) is 0 at x=0. And now differentiate the numerator.
You get N'=$\sin x- x \cos x$.Now as tanx > x . So, $N'\ge 0$ .and thus N increases as x increases.
ie.N +ve . And also denominator(D) is a perfect square, so it is positive.So $f'\ge0$
So the function increases in $[0,2\pi]$.
