<s>[INST] Simplifying and evaluating $\cot 70^\circ+4\cos 70^\circ$ I have to simplify and evaluate this : 
$$\cot 70^\circ+4\cos 70^\circ$$
On evaluating it, the answer comes out to be $1.732$, or $\sqrt 3$ .
I tried to get everything in $\sin$ and $\cos$, but it doesn't go any further. Any hints?
 [/INST] Like Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $,
If $2\sin3x=-1$
$$\cot x+4\cos x=4\cos x-2\sin3x\cot x$$
$$=\dfrac{4\cos x\sin x-2\sin3x\cos x}{\sin x}$$
$$=\dfrac{2\sin2x-(\sin2x+\sin4x)}{\sin x}$$
$$=-2\cos3x\text{ as }\sin x\ne0$$
Here $x=70^\circ$
Generalization: 
$$\sin3x=-\dfrac12\implies3x=180^\circ n+(-1)^n(-30^\circ)\text{ where $n$ is any integer}$$
$\implies x=60^\circ n+(-1)^n(-10^\circ)$  where $n\equiv-1,0,1\pmod3$
$n=-1\implies x=-50^\circ$
$n=0\implies x=-10^\circ$
$n=1\implies x=70^\circ$
</s>