<s>[INST] How do I simplify $\sqrt{3}(\cot 70^\circ + 4 \cos 70^\circ )$? $\sqrt{3}(\cot 70^\circ + 4 \cos 70^\circ ) =$ ? 
The answer is $3$. 
My progress so far:
\begin{align}
\sqrt{3}(\cot 70^\circ+4\cos 70^\circ)&= \sqrt{3}(\tan 20^\circ+4\sin 20^\circ) \\
&= \sqrt{3}(\sin 20^\circ)\left(\frac{1}{\cos 20^\circ}+4\right)\\
&= \sqrt{3}\frac{(\sin 20^\circ)(1+4\cos 20^\circ)}{\cos 20^\circ}\\
&= \sqrt{3}\frac{(\sin 20^\circ+2\sin 40^\circ)}{\cos 20^\circ}\\
\end{align}
 [/INST] This is a repeating use of the sum-to-product identities.
\begin{align}
\frac{\sin 20^\circ + 2\sin 40^\circ}{\cos 20^\circ} &= \frac{2\sin 30^\circ \cos 10^\circ + \sin 40^\circ}{\cos 20^\circ} \\
&= \frac{\cos 10^\circ + \cos 50^\circ}{\cos 20^\circ} \\
&= \frac{2\cos 30^\circ \cos 20^\circ}{\cos 20^\circ} \\
&= \sqrt{3}
\end{align}
so the answer of the original expression is $3$.
</s>