Simplify the expression: $\dfrac{2 \cos^2 \alpha - 1}{4 \tan \left(\dfrac{\pi}{4} - \alpha\right) \sin^2 \left(\dfrac{\pi}{4} + \alpha\right)}$
$\dfrac{2 \cos^2 \alpha - 1}{4 \tan \left(\dfrac{\pi}{4} - \alpha\right) \sin^2 \left(\dfrac{\pi}{4} + \alpha\right)}$
$= \dfrac{\cos 2 \alpha}{4 \left[\dfrac{\tan \dfrac{\pi}{4} - \tan \alpha}{1 + \tan \dfrac{\pi}{4} \tan \alpha}\right] \left[\sin \dfrac{\pi}{4} \cos \alpha + \cos \dfrac{\pi}{4} \sin \alpha\right]^2}$
$= \dfrac{\cos 2 \alpha}{4 \left[\dfrac{1 - \tan \alpha}{1 + \tan \alpha}\right] \left[\dfrac{1}{\sqrt{2}} \cos \alpha + \dfrac{1}{\sqrt{2}} \sin \alpha\right]^2}$
$= \dfrac{\cos 2 \alpha}{4 \left[\dfrac{1 - \dfrac{\sin \alpha}{\cos \alpha}}{1 + \dfrac{\sin \alpha}{\cos \alpha}}\right] \left[\dfrac{\cos^2 \alpha}{2} + \dfrac{\sin^2 \alpha}{2} + 2 \times \dfrac{1}{\sqrt{2}} \cos \alpha \times \dfrac{1}{\sqrt{2}} \sin \alpha\right]}$
$= \dfrac{\cos 2 \alpha}{4 \left[\dfrac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha}\right] \left[\dfrac{1}{2} + \sin \alpha \cos \alpha\right]}$
$= \dfrac{\cos 2 \alpha}{4 \times \dfrac{\left(\cos \alpha - \sin \alpha\right) \left(\cos \alpha + \sin \alpha\right)}{\left(\cos \alpha + \sin \alpha\right)^2} \times \left(\dfrac{1 + 2 \sin \alpha \cos \alpha}{2}\right)}$
$= \dfrac{\cos 2 \alpha \times \left(\cos \alpha + \sin \alpha\right)^2 \times 2}{4 \times \left(\cos^2 \alpha - \sin^2 \alpha\right) \times \left(\sin^2 \alpha + \cos^2 \alpha + 2 \sin \alpha \cos \alpha\right)}$
$= \dfrac{\cos 2 \alpha \times \left(\cos \alpha + \sin \alpha\right)^2}{2 \times \cos 2 \alpha \times \left(\cos \alpha + \sin \alpha\right)^2}$
$= \dfrac{1}{2}$