A trigonometric identity without variables

 We will prove the trigonometric identity

$$\tan10^{\circ}\cdot \tan50^{\circ}\cdot \cot20^{\circ}=\frac{1}{\sqrt{3}} \tag{1}$$

The problem appears at Solutions in GMB 3/2005, number O:1063, page 132. You can find more about this magazine here and here.

Official Solution

      Using that we have

 $\sin50^{\circ}=\cos(90^{\circ}-50^{\circ})=\cos40^{\circ},\;\;\cos10^{\circ}=\sin(90^{\circ}-10^{\circ})=\sin80^{\circ}$

and  $\cos50^{\circ}=\sin(90^{\circ}-50^{\circ})=\sin40^{\circ}$  we obtain

$\tan10^{\circ}\cdot \tan50^{\circ}\cdot \cot20^{\circ}=\frac{\sin10^{\circ}}{\cos10^{\circ}} \cdot \frac{\sin50^{\circ}}{\cos50^{\circ}} \cdot \frac{\cos 20^{\circ}}{\sin20^{\circ}}=\frac{\sin10^{\circ}\cdot \cos40^{\circ}\cdot \cos20^{\circ}}{\sin80^{\circ}\cdot \sin40^{\circ}\cdot \sin20^{\circ} } \tag{2}$

          Up in (2) we have, from  $\sin20^{\circ}=2\cdot \sin10^{\circ}\cdot \cos10^{\circ}$  if we replace  $2\cdot \sin10^{\circ}$ :

$8\cdot \sin10^{\circ}\cdot \cos20^{\circ}\cdot \cos40^{\circ}=\frac{4\cdot \sin20^{\circ}\cdot \cos20^{\circ}\cos40^{\circ}}{ \cos10^{\circ}}=\frac{2\cdot \sin40^{\circ}\cdot \cos40^{\circ}}{\cos10^{\circ}}=\frac{\sin80^{\circ}}{\cos10^{\circ}}=1.$

          Down in (2) we have, using  $2\sin u\sin v=\cos(v-u)-\cos(v+u)$, replacing  $\sin80^{\circ}=\cos10^{\circ}$, and using  $2\cos u \cos v=\cos(v-u)+\cos(v+u)$

$8\cdot \sin20^{\circ}\cdot \sin40^{\circ}\cdot \sin80^{\circ}=4(\cos20^{\circ}-\cos60^{\circ})\cdot \cos10^{\circ}=$

$=4\cdot \cos20^{\circ}\cdot \cos10^{\circ}-4\cdot \frac{1}{2}\cdot \cos10^{\circ}=2(\cos10^{\circ}+\cos30^{\circ})-2\cdot \cos10^{\circ}=$

$=2\cdot \cos10^{\circ}+2\cdot \frac{\sqrt{3}}{2}-2\cdot \cos10^{\circ}=\sqrt{3}$

These values ​​entered in  (2) give the answer.

$\blacksquare$

               Just as "After the battle, everyone is a general", we will also provide a solution.

                    Solution CiP

                Lemma CiP      $t_1=\tan20^{\circ}$  is  one of the roots of the equation

$$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0 \tag{3}$$

               Proof of Lemma

          In the triple angle formula  $\tan 3\theta=\frac{3\tan \theta-\tan^3 \theta}{1-3\tan^2 \theta}$  we substitute

$\theta=20^{\circ},\;\;t=\tan \theta$  and because  $\tan3\theta=\tan 60^{\circ}=\sqrt{3}$, we have

$\frac{3t-t^3}{1-3t^2}=\sqrt{3}\Leftrightarrow\;3t-t^3=\sqrt{3}-3\sqrt{3}\cdot t^2\Leftrightarrow\;t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0$

that is (3).

qed Lemma$\square$

          To prove (1), we successively transcribe it (1)$\;\Leftrightarrow$

$\Leftrightarrow\;\frac{\tan(30^{\circ}-20^{\circ})\cdot \tan(30^{\circ}+20^{\circ})}{\tan20^{\circ}}=\frac{1}{\sqrt{3}}\Leftrightarrow\;\tan(30^{\circ}-20^{\circ})\cdot \tan(30^{\circ}+20^{\circ})=\frac{\tan20^{\circ}}{\sqrt{3}}$

and we have the equivalent problem: 

              Verify that the equation 

$$\tan(30^{\circ}-x)\cdot \tan (30^{\circ}+x)=\frac{\tan x}{\sqrt{3}} \tag{4}$$

               is satisfied by  $x=20^{\circ}$.

The left side of  (4)  is written :

$$\frac{\sin(30^{\circ}-x) \cdot \sin(30^{\circ}+x)}{\cos(30^{\circ}-x)\cdot \cos(30^{\circ}+x)}\;\;\overset{2\sin u\sin v=\cos((v-u)-\cos(v+u)}{\underset{2\cos u\cos v=\cos(v-u)+\cos(v+u)}{=}}$$

$=\frac{\cos2x-\cos60^{\circ}}{\cos2x+\cos60^{\circ}}\;\;\underset{\tan x=t}{=}\;\;\frac{\frac{1-t^2}{1+t^2}-\frac{1}{2}}{\frac{1-t^2}{1+t^2}+\frac{1}{2}}=\frac{2-2t^2-1-t^2}{2-2t^2+1+t^2}=\frac{1-3t^2}{3-t^2}$

so we need to show that the equation  $\frac{1-3t^2}{3-t^2}=\frac{t}{\sqrt{3}}$  is verified by  $t=\tan20^{\circ}$. But the equation with the unknown  $t$ is written equivalently

$\sqrt{3}-3\sqrt{3}\cdot t^2=3\cdot t-t^3\;\Leftrightarrow\;t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0$

which the Lemma shows has the solution  $t=\tan20^{\circ}$.

$\blacksquare$

          Remark CiP  By this we have NOT completely solved the equation  (4) nor what are the other two roots of the equation  (3).

<end Rem>