Identities comprising $tg 10^{\circ},\; tg 50^{\circ},\; tg 70^{\circ}$

           Prove that

$tg^{2} 10^{\circ}+tg^{2} 50^{\circ}+tg^{2} 70^{\circ}=9$.

          Solution CiP

     We have $tg 30^{\circ}=\frac{1}{\sqrt{3}}$ and, applying the formula $tg 3\alpha=\frac{3tg \alpha-tg^{3} \alpha}{1-3tg^{2} \alpha}$,

$\Leftrightarrow \; \frac {3tg 10^{\circ}-tg^{3} 10^{\circ}}{1-3 tg^{2} 10^{\circ}}=\frac{1}{\sqrt{3}}\;\Leftrightarrow \; tg^{3} 10^{\circ}-\sqrt{3} tg^{2} 10^{\circ}-3 tg 10^{\circ}+\frac{1}{\sqrt{3}}=0.$

So $tg 10^{\circ}$ is the root of the third degree equation

(1)               $z^{3}-\sqrt{3} \cdot z^{2}-3 \cdot z +\frac{1}{\sqrt{3}}=0$. 

     The same calculations go for $tg(30^{\circ}+180^{\circ})=\frac{1}{\sqrt{3}}$ and $tg(30^{\circ}+2 \cdot 180^{\circ})=\frac {1}{\sqrt{3}}$, so $tg 70^{\circ}$ and $tg 130^{\circ}=-tg 50^{\circ}$  they are also roots of the equation (1). These are all the roots of the equation (1).

      Now we write Vièta's relations for equation (1) and its roots $tg10^{\circ},\; -tg 50^{\circ}\; tg 70^{\circ}$ - arranging a little -

$$\begin{cases} tg10^{\circ}-tg50^{\circ}+tg70^{\circ}=\sqrt{3}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)\\tg10^{\circ} \cdot tg50^{\circ}+tg50^{\circ} \cdot tg70^{\circ}-tg70^{\circ} \cdot tg10^{\circ}=3\;\;(3)\\tg10^{\circ} \cdot tg 50^{\circ} \cdot tg 70^{\circ}=\frac{1}{\sqrt{3}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(4)\end{cases}$$

 (2)-(4) represent the fundamental relations between the three numbers $tg10^{\circ}, tg50^{\circ},tg70^{\circ}$, from where we can obtain as many symmetrical relations as we want.

    

          From algebraic identity  

$(A+B+C)^{2}=A^{2}+B^{2}+C^{2}+2(A\cdot B+B \cdot C+C \cdot A)$

 
we can express

$tg^{2}10^{\circ}+tg^{2}50^{\circ}+tg^{2}70^{\circ}=(tg10^{\circ}-tg50^{\circ}+tg70^{\circ})^{2}+2\cdot (tg10^{\circ} \cdot tg50^{\circ}+tg50^{\circ} \cdot tg70^{\circ}-tg70^{\circ} \cdot tg10^{\circ})=(\sqrt{3})^{2}+2 \cdot 3=9.$ 

$\blacksquare$

 

This relationship appeared in another context in

 https://www.facebook.com/sanchir.bold.14/posts/3756482811079358

  see the problem on page 221.