Problem with a trigonometric function: $\arctan ( \sin x /(1-\cos x))$

Question

I am studying Abel summability right now, and at a certain point I obtain the following identity: $$ \sum_{k=1}^{\infty}\frac{\sin kx}{k} r^k = \arctan \frac{r\sin x}{1-r\cos x} $$ By previous results, we have that the series $$\sum_{k=1}^{\infty} \frac{\sin kx}{k} $$ converges for all $x\in [0,2\pi ]$, and since Abel summability generalizes convergence, then we have that $$ \lim_{r\to 1^-}\sum_{k=1}^{\infty}\frac{\sin kx}{k} r^k = \sum_{k=1}^{\infty} \frac{\sin kx}{k} = \lim_{r\to 1^-} \arctan \frac{r\sin x}{1-r\cos x} =\arctan \frac{\sin x}{1-\cos x}. $$ It is known that for $x\in [0,2\pi]$, $\sum_{k=1}^{\infty} \frac{\sin kx}{k} =(\pi-x)/2$, however I am not able to prove $$ \arctan \frac{\sin x}{1-\cos x}=\frac{\pi - x}{2}, $$ so any help would be appreciated.

Answer 1

$\tan(\frac{\pi}{2}-\frac{x}{2})=\frac{\sin(\frac{\pi}{2}-\frac{x}{2})}{\cos(\frac{\pi}{2}-\frac{x}{2})}=\frac{\cos(\frac{x}{2})}{\sin(\frac{x}{2})}=\frac{2\sin(\frac{x}{2})\cos(\frac{x}{2})}{2\sin(\frac{x}{2})\sin(\frac{x}{2})}=\frac{\sin x}{1-\cos x}$