Simplify the expression: $\dfrac{1 - \cos 4 \alpha}{\sec^2 2 \alpha - 1} + \dfrac{1 + \cos 4 \alpha}{\text{cosec}^2 2 \alpha - 1}$
$\dfrac{1 - \cos 4 \alpha}{\sec^2 2 \alpha - 1} + \dfrac{1 + \cos 4 \alpha}{\text{cosec}^2 2 \alpha - 1}$
$= \dfrac{2 \sin^2 2 \alpha}{\dfrac{1}{\cos^2 2 \alpha} - 1} + \dfrac{2 \cos^2 2 \alpha}{\dfrac{1}{\sin^2 2 \alpha} - 1}$
$\left[\because \; 1 - \cos 2 \theta = 2 \sin^2 \theta \implies 1 - \cos 4 \theta = 2 \sin^2 2 \theta\right] \;\;\;$ and
$\left[1 + \cos 2 \theta = 2 \cos^2 \theta \implies 1 + \cos 4 \theta = 2 \cos^2 2 \theta\right]$
$= \dfrac{2 \sin^2 2 \alpha \cos^2 2 \alpha}{1 - \cos^2 2 \alpha} + \dfrac{2 \cos^2 2 \alpha \sin^2 2 \alpha}{1 - \sin^2 2 \alpha}$
$= \dfrac{2 \sin^2 2 \alpha \cos^2 2 \alpha}{\sin^2 2 \alpha} + \dfrac{2 \cos^2 2 \alpha \sin^2 2 \alpha}{\cos^2 2 \alpha}$
$= 2 \sin^2 2 \alpha \cos^2 2 \alpha \left(\dfrac{1}{\sin^2 2 \alpha} + \dfrac{1}{\cos^2 2 \alpha}\right)$
$= 2 \sin^2 2 \alpha \cos^2 2 \alpha \left(\dfrac{\cos^2 2 \alpha + \sin^2 2 \alpha}{\sin^2 2 \alpha \cos^2 2 \alpha}\right)$
$= 2 \times 1 = 2$