The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$ The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$=_____
My approach is as follow
Already this question is solved Maximum value of argument but I would like to slve by my approach which is illustrated below
$z = \cos \theta  + i\sin \theta ;\theta  \notin \left\{ 0 \right\}$ because $z\ne1$
$\left| {\operatorname{Arg}\left( {\frac{1}{{1 - \cos \theta  - i\sin \theta }}} \right)} \right|$
$\left| {\operatorname{Arg}\left( {\frac{1}{{2{{\sin }^2}\frac{\theta }{2} - 2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right)} \right|$
$\left| {\operatorname{Arg}\left( {\frac{1}{{2\sin \frac{\theta }{2}\left( {\sin \frac{\theta }{2} - i\cos \frac{\theta }{2}} \right)}}} \right)} \right| \Rightarrow \left| {\operatorname{Arg}\left( {\frac{{\left( {\sin \frac{\theta }{2} + i\cos \frac{\theta }{2}} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$
$\left| {\operatorname{Arg}\left( {\frac{{\left( {\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) + i\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$
How do we get the required answer which if $\frac{\pi}{2}$

Hint:
Arg$\left(\dfrac12+i\cdot\dfrac{\cot\dfrac\theta2}2\right)=$Arg$\left(\cot\dfrac\theta2\right)=$Arg$\left(\tan\left(\dfrac{\pi-\theta}2\right)\right)$
Use this
