Transform the following expression into a product:
$\sec \alpha - \cos \alpha + \sec 60^\circ \cos 2 \alpha \sin 3 \alpha - \sin 5 \alpha$
Given: $\;$ $\sec \alpha - \cos \alpha + \sec 60^\circ \cos 2 \alpha \sin 3 \alpha - \sin 5 \alpha$
$= \dfrac{1}{\cos \alpha - \cos \alpha + 2 \cos 2 \alpha \sin 3 \alpha - \sin 5 \alpha}$
$= \dfrac{1}{\cos \alpha} - \cos \alpha + \sin \left(3 \alpha - 2 \alpha\right) + \sin \left(3 \alpha + 2 \alpha\right) - \sin 5 \alpha$
$\left[\because \; 2 \sin A \cos B = \sin \left(A - B\right) + \sin \left(A + B\right)\right]$
$= \dfrac{1}{\cos \alpha} - \cos \alpha + \sin \alpha + \sin 5 \alpha - \sin 5 \alpha$
$= \dfrac{1}{\cos \alpha} - \cos \alpha + \sin \alpha$
$= \dfrac{1 - \cos^2 \alpha}{\cos \alpha} + \sin \alpha$
$= \dfrac{\sin^2 \alpha}{\cos \alpha} + \sin \alpha$
$= \sin \alpha \left(\dfrac{\sin \alpha}{\cos \alpha} + 1\right)$
$= \dfrac{\sin \alpha}{\cos \alpha} \left(\sin \alpha + \cos \alpha\right)$
$= \dfrac{\sin \alpha}{\cos \alpha} \times \sqrt{2} \times \left(\dfrac{1}{\sqrt{2}} \sin \alpha + \dfrac{1}{\sqrt{2}} \cos \alpha\right)$
$= \sqrt{2} \tan \alpha \left(\sin \dfrac{\pi}{4} \cos \alpha + \cos \dfrac{\pi}{4} \sin \alpha\right)$
$= \sqrt{2} \tan \alpha \sin \left(\dfrac{\pi}{4} + \alpha\right)$ $\;\; \left[\because \; \sin \left(A + B\right) = \sin A \cos B + \cos A \sin B\right]$