Show that

$ \left(1-e^{i\theta}\right)^{-\frac{1}{2}} \left(1-e^{-i\theta}\right)^{-\frac{1}{2}}=\left(1 \mathrm{cosec} \left(\frac{\theta}{2}\right)\right)^{\frac{1}{2}} $

$Let \ y=(1-e^{i\theta})^{-\frac {1}{2}} \cdots \ \cdots (A) $ $y^2 \left [ \left ( 1 - e^{i\theta} \right )^{-\frac {1}{2}} + \left ( 1 - e^{-i \theta} \right )^{-\frac {1}{2}} \right ]^2 $ $= \left ( 1 - e^{i\theta} \right )^{-1} + \left ( 1-e^{-i\theta} \right )^{-1} + 2 \left ( 1 - e^{i \theta} \right )^{-\frac {1}{2}} \left ( 1-e^{i \theta} \right )^{-\frac {1}{2}} $ $= \dfrac {\left ( 1 - e^{i\theta} \right )+ \left ( 1-e^{-i\theta} \right )}{ \left ( 1 - e^{i \theta} \right ) \left ( 1 - e^{- i \theta} \right )}+2 \left [ \left ( 1-e^{i\theta} \right ) \left ( 1 - e^{-i\theta} \right ) \right ]^{- \frac {1}{2}}$ $Now, \ \left ( 1-e^{i\theta} \right ) + \left ( 1-e^{-i \theta} \right )= 1 -e^{i \theta}- e^{-i\theta}+ 1=2-2 \left ( \dfrac {e^{i\theta}+ e^{-i\theta}} {2} \right ) =2 -2 \cos \theta$ $and \ (1-e^{i \theta})+ (1-e^{-i\theta})= 1 -e^{i \theta}- e^{-i\theta}+ 1 =2-2 \left ( \dfrac {e^{i \theta}+ e^{-i\theta}} {2} \right ) = 2 -2 \cos \theta$ $y^2 = \dfrac {2-2 \cos \theta}{2 - 2 \cos 2}+ 2 \left [ 2-2 \cos \theta \right ]^{-\frac {1}{2}} = 1 + \dfrac {2}{\sqrt{2 (1-\cos \theta)}}$ $y^2 = 1 + \dfrac {2}{\sqrt{2 \left (2 \sin^2 \frac {\theta}{2} \right )}} \ \ \ {\cos 2x=1 -2 \sin^2 x } $ $y^2 1 + \dfrac {2}{2 \sin \left ( \frac {\theta}{2} \right )} = 1 + cosec \left ( \dfrac {\theta} {2} \right ) $ $\therefore y= \sqrt{1+cosec \left ( \dfrac { \theta} {2} \right )} \ \cdots \ \cdots (B)$ From (A) and (B), $\left ( 1 - e^{i\theta} \right )^{-\frac {1}{2}} + \left ( 1 - e^{-i \theta} \right )^{-\frac {1}{2}} = \left ( 1 + cosec \left ( \dfrac {\theta} {2} \right ) \right )^{\frac {1}{2}}$