Transform the following expression into a product:
$\tan^2 \alpha - \tan^2 \beta - \dfrac{1}{2} \sin \left(\alpha - \beta\right) \sec^2 \alpha \sec^2 \beta$
Given: $\;$ $\tan^2 \alpha - \tan^2 \beta - \dfrac{1}{2} \sin \left(\alpha - \beta\right) \sec^2 \alpha \sec^2 \beta$
$= \dfrac{\sin^2 \alpha}{\cos^2 \alpha} - \dfrac{\sin^2 \beta}{\cos^2 \beta} - \dfrac{1}{2} \sin \left(\alpha - \beta\right) \sec^2 \alpha \sec^2 \beta$
$= \dfrac{\sin^2 \alpha \cos^2 \beta - \sin^2 \beta \cos^2 \alpha}{\cos^2 \alpha \cos^2 \beta} - \dfrac{1}{2} \sin \left(\alpha - \beta\right) \sec^2 \alpha \sec^2 \beta$
$= \sec^2 \alpha \sec^2 \beta \left[\sin^2 \alpha \cos^2 \beta - \cos^2 \alpha \sin^2 \beta\right] - \dfrac{1}{2} \sin \left(\alpha - \beta\right) \sec^2 \alpha \sec^2 \beta$
$= \sec^2 \alpha \sec^2 \beta \left[\left(\sin \alpha \cos \beta + \cos \alpha \sin \beta\right) \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right)\right]$
$\hspace{3cm} - \sec^2 \alpha \sec^2 \beta \times \dfrac{1}{2} \sin \left(\alpha - \beta\right)$
$= \sec^2 \alpha \sec^2 \beta \left[\sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta\right) - \dfrac{1}{2} \sin \left(\alpha - \beta\right)\right]$
$\left\{\because \; \sin \left(A \pm B\right) = \sin A \cos B \pm \cos A \sin B\right\}$
$= \sec^2 \alpha \sec^2 \beta \left[\sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta\right) - \sin \left(\dfrac{\pi}{6}\right) \sin \left(\alpha - \beta\right)\right]$
$\left\{\because \; \sin \left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}\right\}$
$= \sec^2 \alpha \sec^2 \beta \sin \left(\alpha - \beta\right) \left[\sin \left(\alpha + \beta\right) - \sin \left(\dfrac{\pi}{6}\right)\right]$
$= \sec^2 \alpha \sec^2 \beta \sin \left(\alpha - \beta\right) \times 2 \sin \left(\dfrac{\alpha + \beta - \dfrac{\pi}{6}}{2}\right) \cos \left(\dfrac{\alpha + \beta + \dfrac{\pi}{6}}{2}\right)$
$\left\{\because \; \sin A - \sin B = 2 \sin \left(\dfrac{A - B}{2}\right) \cos \left(\dfrac{A + B}{2}\right)\right\}$