<s>[INST] Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $ How to find the value of
$$\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$$
manually ?
 [/INST] Let $a\cos\theta+b\cot\theta=c$
$\implies a\cos\theta=c-b\cot\theta=\frac{c\sin\theta-b\cos\theta}{\sin\theta}$
$\implies a\cos\theta\sin\theta=c\sin\theta-b\cos\theta$
Putting $c=r\cos\alpha,b=r\sin\alpha$ where $r>0$
Squaring & adding we get $r^2=c^2+b^2\implies r=+\sqrt{b^2+c^2}$ and $\frac{\sin\alpha}b=\frac{\cos\alpha}c=\frac1r=\frac1{\sqrt{b^2+c^2}}$
$\implies c\sin\theta-b\cos\theta=\sqrt{b^2+c^2}\sin(\theta-\alpha)$ and $\cos\theta\sin\theta=\frac{\sin2\theta}2$
$$a\sin2\theta=2\sqrt{b^2+c^2}\sin(\theta-\alpha)$$
Now, the solution of $P\sin x= Q\sin A $ is in general intractable unless  $P=0$ or $Q=0$ or $P=\pm Q\ne0$
Here 
$1:$ if $a=0,$ the problem can be solved easily.
$2:$ if $b^2+c^2=0\implies b=c=0$ ( as $b,c$ are real), the problem can be solved easily.
$3:$  So, for non-trivial cases, either $a=2\sqrt{b^2+c^2}$ or $a=-2\sqrt{b^2+c^2}$
$$\begin{array}{|c|c|c|} 
 \hline \text{ Case } & a=2\sqrt{b^2+c^2} & a=-2\sqrt{b^2+c^2}  \\ 
 \hline \text{ Comparison} & \sin2\theta=\sin(\theta-\alpha)
& \sin2\theta=-\sin(\theta-\alpha)=\sin(\alpha-\theta),\text{ as }\sin(-x)=-\sin x   \\
\hline \text{General Solution} & 2\theta=n180^\circ+(-1)^n(\theta-\alpha)\text{  where }n\text{ is any integer } & 2\theta=n180^\circ+(-1)^n(\alpha-\theta)\text{  where }n\text{ is any integer } \\
 \hline n=2m & \alpha=m360^\circ-\theta\equiv-\theta\pmod{360^\circ} &  \alpha=3\theta-m360^\circ\equiv3\theta \\ 
 \hline n=2m+1& \alpha=3\theta-(2m+1)180^\circ\equiv 3\theta-180^\circ &  \alpha=(2m+1)180^\circ-\theta\equiv180^\circ-\theta  \\
  \hline
  \end{array} $$
Here $a=-4,b=\sqrt3,\theta=20^\circ, $
So, $a=-2\sqrt{b^2+c^2}$ as $\sqrt{b^2+c^2} > 0$
$\implies\sqrt{b^2+c^2}=-\frac a2=2, \alpha=3\theta=60^\circ$ or $\alpha=180^\circ-\theta=160^\circ$
So, $\sin\alpha=\frac b{\sqrt{b^2+c^2}}=\frac{\sqrt3}2\implies \alpha=60^\circ$
So, $c=\cos\alpha\cdot \sqrt{b^2+c^2}=\frac12\cdot 2=1$
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