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 ?
 A: 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, $c=\cos\alpha\cdot \sqrt{b^2+c^2}=\frac12\cdot 2=1$
A: If $\cos3\theta=\frac12,$
$$\tan3\theta\cot\theta-4\cos\theta$$
$$=\frac{\sin3\theta\cos\theta}{\cos3\theta\sin\theta}-4\cos\theta$$
$$=\frac{\sin3\theta\cos\theta-4\cos3\theta\sin\theta\cos\theta}{\cos3\theta\sin\theta}$$
$$=\frac{\sin3\theta\cos\theta-\sin2\theta}{\frac12\sin\theta}\text { as }\sin2x=2\sin x\cos x\text{ and } \cos3\theta=\frac12$$
$$=\frac{\sin4\theta+\sin2\theta-2\sin2\theta}{\sin\theta}\text { applying } 2\sin A\cos B=\sin(A+B)+\sin(A-B)$$
$$=\frac{\sin4\theta-\sin2\theta}{\sin\theta}$$
$$=\frac{2\sin\theta\cos3\theta}{\sin\theta}\text { applying } \sin 2C-\sin 2D=2\sin(C-D)\cos(C+D)$$
$$=2\cos3\theta=1$$
Now, $$\cos3\theta=\frac12=\cos60^\circ\implies 3\theta=2n180^\circ\pm60^\circ\text{ where } n \text{ is any integer}$$
So, $\theta=(6n\pm1)20^\circ$  where $n$ is any integer
A: Like Trigonometry Simplification,
$$\frac{2\sin60^\circ\cdot\cos20^\circ-2(2\sin20^\circ\cos20^\circ)}{\sin20^\circ}$$
Using Werner Formula we get,
$$\frac{\sin80^\circ+\sin40^\circ-2\sin40^\circ}{\sin20^\circ}$$
Using Prosthaphaeresis Formula,  $\sin80^\circ-\sin40^\circ=2\sin20^\circ\cos60^\circ$
A: $$\displaystyle \sin(60^{\circ}-20^{\circ}) = \sin 40^{\circ} = 2 \sin 20^{\circ} \cos 20^{\circ}$$
$$\displaystyle \frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ} = 2 \sin 20^{\circ} \cos 20^{\circ}$$
$$\displaystyle \frac{\sqrt{3}}{2} \cos 20^{\circ}  - 2 \sin 20^{\circ} \cos 20^{\circ}  = \frac{1}{2} \sin 20^{\circ}$$
Multiply by $\displaystyle \frac{2}{\sin 20^{\circ}}$
$$\displaystyle \sqrt{3} \cot 20^{\circ} - 4 \cos 20^{\circ} = 1$$
A: For $\sin x\ne0,$
$$2\sin3x\cot x-4\cos x$$
$$=\dfrac{2\sin3x\cos x-2(2\cos x\sin x)}{\sin x}$$
$$=\dfrac{\sin4x+\sin2x-2\sin2x}{\sin x}$$
$$=\dfrac{\sin4x-\sin2x}{\sin x}$$
$$=2\cos3x$$
Now set $\   2\sin3x=\dfrac{\sqrt3}2=\sin60^\circ\implies 3x=n180^\circ+(-1)^n60^\circ\text{ where } n \text{ is any integer}$
So, $x=60^\circ n+(-1)^n20^\circ$  where $n\equiv-1,0,1\pmod3$
Here $n=0$
