Prove the identity: $\dfrac{\sqrt{1 + \cos \alpha} + \sqrt{1 - \cos \alpha}}{\sqrt{1 + \cos \alpha} - \sqrt{1 - \cos \alpha}} = \cot \left(\dfrac{\alpha}{2} + \dfrac{\pi}{4}\right)$, $\;\;\;$ $\pi < \alpha < 2 \pi$
$\begin{aligned}
LHS & = \dfrac{\sqrt{1 + \cos \alpha} + \sqrt{1 - \cos \alpha}}{\sqrt{1 + \cos \alpha} - \sqrt{1 - \cos \alpha}} \\\\
& = \dfrac{\left(\sqrt{1 + \cos \alpha} + \sqrt{1 - \cos \alpha}\right)^2}{\left(\sqrt{1 + \cos \alpha} - \sqrt{1 - \cos \alpha}\right) \left(\sqrt{1 + \cos \alpha} + \sqrt{1 - \cos \alpha}\right)} \\\\
& = \dfrac{1 + \cos \alpha + 1 - \cos \alpha - 2 \sqrt{\left(1 + \cos \alpha\right) \left(1 - \cos \alpha\right)}}{1 + \cos \alpha - \left(1 - \cos \alpha\right)} \\\\
& = \dfrac{2 - 2 \sqrt{1 - \cos^2 \alpha}}{2 \cos \alpha} \\\\
& = \dfrac{1 - \sqrt{\sin^2 \alpha}}{\cos \alpha} \\\\
& = \dfrac{1 - \sin \alpha}{\cos \alpha} \\\\
& = \dfrac{\sin^2 \left(\dfrac{\alpha}{2}\right) + \cos^2 \left(\dfrac{\alpha}{2}\right) - 2 \sin \left(\dfrac{\alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right)}{\cos^2 \left(\dfrac{\alpha}{2}\right) - \sin^2 \left(\dfrac{\alpha}{2}\right)} \\\\
& = \dfrac{\left[\cos \left(\dfrac{\alpha}{2}\right) - \sin \left(\dfrac{\alpha}{2}\right)\right]^2}{\left[\cos \left(\dfrac{\alpha}{2}\right) + \sin \left(\dfrac{\alpha}{2}\right)\right] \left[\cos \left(\dfrac{\alpha}{2}\right) - \sin \left(\dfrac{\alpha}{2}\right)\right]} \\\\
& = \dfrac{\cos \left(\dfrac{\alpha}{2}\right) - \sin \left(\dfrac{\alpha}{2}\right)}{\cos \left(\dfrac{\alpha}{2}\right) + \sin \left(\dfrac{\alpha}{2}\right)} \\\\
& = \dfrac{1 - \dfrac{\sin \left(\dfrac{\alpha}{2}\right)}{\cos \left(\dfrac{\alpha}{2}\right)}}{1 + \dfrac{\sin \left(\dfrac{\alpha}{2}\right)}{\cos \left(\dfrac{\alpha}{2}\right)}} \\\\
& = \dfrac{1 - \tan \left(\dfrac{\alpha}{2}\right)}{1 + \tan \left(\dfrac{\alpha}{2}\right)} \\\\
& = \dfrac{1 - \tan \left(\dfrac{\alpha}{2}\right) \tan \left(\dfrac{\pi}{4}\right)}{\tan \left(\dfrac{\alpha}{2}\right) + \tan \left(\dfrac{\pi}{4}\right)} \\\\
& = \cot \left(\dfrac{\alpha}{2} + \dfrac{\pi}{4}\right) = RHS
\end{aligned}$
Hence proved.