Question Number 35610 by abdo mathsup 649 cc last updated on 21/May/18
$$\left.{let}\:{give}\:{x}\in\right]\mathrm{0},\mathrm{2}\pi\left[\:\:{and}\:{a}\:\in{R},{b}\in\:{R}\right. \\ $$$${prove}\:{that}\:\:\frac{\pi−{x}}{\mathrm{2}}\:=\:{arctan}\left(\frac{{sinx}}{\mathrm{1}−{cosx}}\right) \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\mid{arctan}\left({a}\right)−{arctan}\left({b}\right)\mid\leqslant\mid{a}−{b}\mid \\ $$$$\left.\mathrm{3}\left.\right){let}\theta\:\in\right]\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\:\:,\:{x}\:\in\left[\theta,\mathrm{2}\pi−\theta\right]\:,\:{r}\in\left[\mathrm{0},\mathrm{1}\left[\:{prove}\:{that}\right.\right.\right. \\ $$$$\mid\varphi\left({x},{r}\right)\:−\frac{\pi−{x}}{\mathrm{2}}\mid\leqslant\:\:\frac{\mathrm{1}−{r}}{\left(\mathrm{1}−{cos}\theta\right)^{\mathrm{2}} } \\ $$