Any better way to find $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$ 

Find the value of  $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$


My Method:
I used the following Identities:
\begin{aligned}
& \sin A \cos B-\cos A \sin B=\sin (A-B) \\
& 2 \sin A \sin B=\cos (A-B)-\cos (A+B) \\
& \cos (2 A)=2 \cos ^2 A-1 \\
& \cos (3 A)=4 \cos ^3 A-3 \cos A \\
& \sin (2 A)=2 \sin A \cos A \\
& 2 \sin A \cos B=\sin (A+B)+\sin (A-B) \\
&
\end{aligned}
$$\begin{aligned}
 S_1 & =\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)=\cot(10^{\circ})-\cot \left(50^{\circ}\right)+\tan(20^{\circ}) \\
\\
 \Rightarrow S_1 & =\frac{\cos \left(10^{\circ}\right)}{\sin \left(10^{\circ}\right)}-\frac{\cos \left(50^{\circ}\right)}{\sin \left(50^{\circ}\right)}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\
\\
\\
 \Rightarrow S_1 & =\frac{2 \sin \left(40^{\circ}\right)}{2 \sin \left(10^{\circ}\right) \sin \left(50^{\circ}\right)}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\
\\
 \Rightarrow S_1 & =\frac{2 \sin \left(40^{\circ}\right)}{\cos \left(40^{\circ}\right)-\frac{1}{2}}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\ \\
 \Rightarrow S_1 & =\frac{4 \sin 40^{\circ}}{4 \cos ^2\left(20^{\circ}\right)-3}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)}=\frac{6 \sin \left(40^{\circ}\right) \cos \left(20^{\circ}\right)-3 \sin \left(20^{\circ}\right)}{0.5}
\end{aligned}$$
$$\begin{aligned}
 & \Rightarrow S_1=\frac{3}{0.5}\left(2 \sin \left(40^{\circ}\right) \cos \left(20^{\circ}\right)-\sin \left(20^{\circ}\right)\right) \\ \\
 & \Rightarrow S_1=6\left(\sin \left(60^{\circ}\right)\right)=3 \sqrt{3}
\end{aligned}$$
 A: I think that I got a different way
Another identity
$$\tan(x+y+z) = \frac{\tan x + \tan y + \tan z - \tan x \tan y \tan z}{1 - \tan y \tan z - \tan x \tan y - \tan x \tan z}$$
Now let $x = 20^{\circ}, y = -40^{\circ}, z=80^{\circ}$, so that $x+y+z = 60^{\circ}$
Let's do some math, using your identities in your post
\begin{align*}
&\tan x \tan (-y) \tan z \\
&= \frac{\sin (20^{\circ})\sin(40^{\circ})\sin(80^{\circ})}{\cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})} \\
&= \frac{\sin(40^{\circ})\sin(80^{\circ})\sin(160^{\circ})}{\cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})} \\
&= 8\sin(20^{\circ})\sin(40^{\circ})\sin(80^{\circ}) \\
&= 4(\cos(20^{\circ}) - \cos(60^{\circ}))\sin(80^{\circ}) \\
&= 4\sin(70^{\circ})\sin(80^{\circ}) - 2\sin(80^{\circ}) \\
&= 2(\cos(10^{\circ}) - \cos(150^{\circ})) - 2\sin(80^{\circ}) \\
&= 2\cos(30^{\circ}) \\
&= \sqrt{3}
\end{align*}
Using the identity
$$
\tan (a-b) = \frac{\tan a - \tan b}{1+\tan a \tan b}
$$
we obtain
\begin{align*}
&\tan x \tan z + \tan x \tan y + \tan x \tan z \\
=& -3 + (1 + \tan x \tan z) + (1 + \tan x \tan y) + (1 + \tan z \tan y) \\
=& -3 
+ \frac{\tan (20^{\circ}) - \tan (80^{\circ})}{\tan(-60^{\circ})}+ \frac{\tan(20^{\circ}) + \tan(40^{\circ})}{\tan(60^{\circ})} + \frac{\tan(80^{\circ}) + \tan(40^{\circ})}{\tan(120^{\circ})} \\
=& -3
\end{align*}
By the identity at the beginning, we obtain
\begin{align*}
\tan(60^{\circ}) &= \frac{\tan x + \tan y + \tan z + \sqrt{3}}{1 + 3} \\
4\sqrt{3} &= \tan x + \tan y + \tan z + \sqrt{3} \\
3\sqrt{3} &= \tan(20^{\circ}) - \tan(40^{\circ}) + \tan(80^{\circ})
\end{align*}
Thus, $\cot(10^{\circ}) + \cot(70^{\circ}) - \cot(50^{\circ}) = 3\sqrt{3}$
A: Since $$\tan{x}+\tan\left(60^{\circ}+x\right)+\tan\left(120^{\circ}+x\right)=$$
$$=\tan{x}+\frac{\sqrt3+\tan{x}}{1-\sqrt3\tan{x}}+\frac{-\sqrt3+\tan{x}}{1+\sqrt3\tan{x}}=3\tan3x,$$ we obtain:
$$\cot10^{\circ}+\cot70^{\circ}-\cot50^{\circ}=\tan80^{\circ}+\tan20^{\circ}+\tan140^{\circ}=3\tan(3\cdot20^{\circ})=3\sqrt3.$$
A: Yet an other way to proceed, which arguably explains why we get a "nice value" for the sum. Let $c_1,c_2,c_3$ be the three different values of the cotangent function computed in $x_1,x_2,x_3$, which are respectively
$10^\circ$, $70^\circ$, $130^\circ$. These angles are of the shape $10^\circ+k\cdot 60^\circ$, so taking the cotangent of the triple value we get
$\cot 30^\circ=\sqrt3$. There is a formula for $\cot (3x)$ in terms of $c:=\cot x$ obtained from $\displaystyle\cot(x+y)=(\cot x\cot y-1)/(\cot x+\cot y)$, namely
$$
\cot(3x)=\frac{c^3-3c}{3c^2-1}=:f(c)\ .
$$
Plugging in $c_1,c_2,c_3$ into $f$ we thus obtain every time $\cot 30 ^\circ=\sqrt 3$. So $c_1,c_2,c_3$ are three (thus all) roots of the polynomial equation:
$$
c^3-3c=\sqrt3(3c^2-1)\ ,\qquad\text{ i.e. }\qquad
c^3-3\sqrt 3c^2-3c+\sqrt 3=0\ .
$$
From here we obtain immediately relations like:
$$
\begin{aligned}
\color{blue}{3\sqrt 3} &= \color{blue}{c_1+c_2+c_3}\ ,\\
-3 &= c_1c_2 + c_3c_3+c_3c_1\ ,\\
-\sqrt 3 &= c_1c_2c_3\ ,\\[3mm]
33 &=c_1^2+c_2^2+c_3^2\ ,\\
105\sqrt3 &=c_1^3+c_2^3+c_3^3\ .
\end{aligned}
$$
The blue relation is the needed one.
$\square$

Computer check:
sage: c1, c2, c3 = cot(pi/18), cot(7*pi/18), cot(13*pi/18)
....: (c1 + c2 + c3).minpoly()
....: (c1 + c2 + c3).numerical_approx()
x^2 - 27
5.19615242270663

So $c_1+c_2+c_3$ is the positive root of the polynomial $x^2-27$.
