The problem can also be interepreted as to simplify
$\begin{align*}(\sqrt{3}\sec 36^{\circ}+\tan6^{\circ})\tan 24^{\circ}&=(\dfrac{\sqrt{3}}{\cos 36^{\circ}}+\tan6^{\circ})\tan 24^{\circ}\\&=\dfrac{\sqrt{3}\tan 24^{\circ}}{\cos 36^{\circ}}+\tan6^{\circ}\tan 24^{\circ}\end{align*}$
It's really hard to "predict" what would be the next best step to simplify the given expression, but my plan of attack is to find an expression that relates between $\tan 24^{\circ}$ and $\cos 36^{\circ}$.
From the identity [TABLE="class: grid, width: 800"]
[TR]
[TD]$\tan (90^{\circ}-y)=\dfrac{1}{\tan y}$[/TD]
[TD]and[/TD]
[TD]$\tan x \tan (60^{\circ}-x) \tan (60^{\circ}-x)=\tan 3x$, we see that if we set $x=24^{\circ}$, we have[/TD]
[/TR]
[/TABLE]
$\tan 24^{\circ} \tan 36^{\circ} \tan 84^{\circ}=\tan 72^{\circ}$
$\tan 24^{\circ} \tan 36^{\circ} \dfrac{1}{\tan 6^{\circ}}=\dfrac{1}{\tan 18^{\circ}}$
$\tan 24^{\circ} \tan 36^{\circ} \tan 18^{\circ}=\tan 6^{\circ}$
$\tan 24^{\circ} \dfrac{\sin 36^{\circ}}{\cos 36^{\circ}} \dfrac{\sin 18^{\circ}}{\cos 18^{\circ}}=\tan 6^{\circ}$
$\tan 24^{\circ} \dfrac{2\sin 18^{\circ}\cancel{\cos 18^{\circ}}}{\cos 36^{\circ}} \dfrac{\sin 18^{\circ}}{\cancel{\cos 18^{\circ}}}=\tan 6^{\circ}$
$\tan 24^{\circ}(2\sin^2 18^{\circ})=\tan 6^{\circ}\cos 36^{\circ}$
$\tan 24^{\circ}(1-\cos 36^{\circ})=\tan 6^{\circ}\cos 36^{\circ}$
$\dfrac{\sqrt{3}\tan 24^{\circ}}{\cos 36^{\circ}}=1-(\tan 6^{\circ}\tan 24^{\circ})$
$\begin{align*}\therefore (\sqrt{3}\sec 36^{\circ}+\tan6^{\circ})\tan 24^{\circ}&=\dfrac{\sqrt{3}\tan 24^{\circ}}{\cos 36^{\circ}}+\tan6^{\circ}\tan 24^{\circ}\\&=1-(\tan 6^{\circ}\tan 24^{\circ})+\tan6^{\circ}\tan 24^{\circ}\\&=1\end{align*}$
