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}
 A: $$\cot(70)+4\cos(70) = \frac{\cos(70)}{\sin(70)} + 4\cos(70) =\frac{\cos(70)+4\cos(70)\sin(70)}{\sin(70)} = \frac{\cos(70)+2\sin(140)}{\sin(70)}  = \frac{(\cos(70)+\sin(140))+\sin(140)}{\sin(70)} = \frac{(\sin(20)+\sin(140))+\sin(140)}{\sin(70)} = \frac{2\sin(80)\cos(60)+\sin(140)}{\sin(70)} = \frac{2\sin(110)\cos(30)}{\sin(70)} = 2 * \frac{\sqrt3}{2} = \sqrt 3$$
Note $\sin(110)=\sin(70)$ at the end.
$\cos(60)=\frac{1}{2}$ and 
$\cos(30)=\frac{\sqrt3}{2}$
Identities used: $\sin(90-x)=\cos(x)$
$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$
$\sin(2x)=2\sin(x)\cos(x)$
$\sin(a)+\sin(b)=2\sin(1/2(a+b))\cos(1/2(a-b))$
A: To continue from your method:
$$\sqrt{3}\frac{(sin20+2sin40)}{cos20}=2\left(\frac{\frac{\sqrt{3}}{2}sin20+2\frac{\sqrt{3}}{2}sin40}{cos20}\right)$$  Using $\frac{\sqrt{3}}{2}=cos30$ we get
$$2\left(\frac{\frac{\sqrt{3}}{2}sin20+2\frac{\sqrt{3}}{2}sin40}{cos20}\right)=\left(\frac{2cos30sin20+2 \times 2cos30sin40}{cos20}\right)=\frac{sin50-sin10+2(sin70+sin10)}{cos20}=\frac{sin50+sin10+2sin70}{cos20} $$
But $sin50+sin10=cos20$ and $sin70=cos20$ so 
$$\frac{sin50+sin10+2sin70}{cos20}=\frac{cos20+2cos20}{cos20}=3 $$
A: 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 $3$.
