# 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}

• math.stackexchange.com/questions/1640343/… Commented Dec 30, 2022 at 16:07
• @labbhattacharjee Wow. I followed the analysis. Hard to believe that an undergraduate would be expected to solve this one. Commented Dec 30, 2022 at 16:44

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: $$\cot10^{\circ}+\cot70^{\circ}-\cot50^{\circ}=\tan80^{\circ}+\tan20^{\circ}+\tan140^{\circ}=3\tan(3\cdot20^{\circ})=3\sqrt3.$$

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 $$x+y+z = 60^{\circ}$$

\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, $$\cot(10^{\circ}) + \cot(70^{\circ}) - \cot(50^{\circ}) = 3\sqrt{3}$$

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 $$x^2-27$$.