$Let \ x= sech^{-1} (\sin \theta )$ $sech \ x=\sin \theta = \dfrac {1}{\cos h \ x} \ \dfrac {2}{e^x+ e^{-x}} = \sin \theta \ e^x+e^{-x}= \dfrac {2}{\sin \theta} \ $ Multiplying ex on both sides, $e^{2x}- \dfrac {2e^x}{\sin \theta}+1=0\$

$This \ is \ a \ quadratic \ \in e^x , \ applying \ quadratic \ formula, \ $

$e^x= \dfrac {- \left ( \dfrac {-2}{\sin \theta}\right )\pm \sqrt{\left ( \dfrac {-2}{\sin \theta} \right )^2 -4}} {2}$

$= \dfrac {-2 \left (- \dfrac {1}{\sin \theta} \right )\pm 2 \sqrt{\left ( - \dfrac {1}{\sin \theta} \right )^2-1}} {2} \\ $

$ = \dfrac {1}{\sin \theta} \pm \sqrt{ \dfrac {1}{\sin^2 \theta}-1}$

$= \dfrac {1}{\sin \theta} \pm \sqrt{\dfrac {1-\sin^2 \theta}{\sin^2 \theta}} \\$ $ = \dfrac {1}{\sin \theta}\pm \sqrt{\dfrac {\cos^2 \theta}{\sin^2 \theta}}= \dfrac {1}{\sin \theta}\pm \dfrac {\cos \theta}{\sin \theta}= \dfrac {1\pm \cos \theta}{\sin \theta}$ Since ex can never be nagative, considering  only positive root, $e^x = \dfrac {1+ \cos \theta}{\sin \theta}= \dfrac {2 \cos^2 \frac {\theta}{2}}{2 \sin \frac {\theta}{2}\cos \frac {\theta}{2}}= \dfrac {\cos \frac {\theta}{2}}{\sin \frac {\theta}{2}}= \cot \dfrac {\theta}{2}$ Taking log on both sides, $\log e^x =x = \log \cot \dfrac {\theta}{2}$ $\therefore sech^{-1}(\sin \theta)= \log \cot \dfrac {\theta}{2}$