How do we prove that :$\tan^2(10)+\tan^2(50)+\tan^2(70) =9$

Question

Prove : $\tan^2(10) + \tan^2(50) + \tan^2(70) =9$

my attempt

Let $\text{t} :=\tan(10)$

$$\tan^2(10) + \tan^2(50) + \tan^2(70) = \tan^2(10) + \tan^2(60-10) + \tan^2(60+10)=t^2 + \left({\frac{\sqrt{3}-t}{1+\sqrt{3}t}}\right)^2+\left({\frac{\sqrt{3}+t}{1-\sqrt{3}t}}\right)^2=\frac{9t^6+45t^2+6}{(1-3t^2)^2}$$

and therfore :

$$\tan^2(10) + \tan^2(50) + \tan^2(70) =\frac{9\tan^6(10)+45\tan^2(10)+6}{(1-3\tan^2(10))^2}$$

I need help completing the proof or give another way and I appreciate everyone's interest

Answer 1

Using your notation, we have $$ \tan(20)=\frac{2\tan(10)}{1-\tan^2(10)}=\frac{2t}{1-t^2} \tag{1} $$ and $$ \tan(30)=\frac{\tan(20)+\tan(10)}{1-\tan(20)\tan(10)}=\frac{\frac{2t}{1-t^2}+t}{1-\frac{2t^2}{1-t^2}}=\frac{3t-t^3}{1-3t^2}. \tag{2} $$ Squaring $(2)$ and simplifying, we obtain $$ \frac{t^6-6t^4+9t^2}{1-6t^2+9t^4}=\tan^2(30)=\frac{1}{3}\implies t^6=9t^4-11t^2+\frac{1}{3}. \tag{3} $$ We can now use $(3)$ to rewrite the RHS of your last equation as $$ \frac{9\left(9t^4-11t^2+\frac{1}{3}\right)+45t^2+6}{(1-3t^2)^2}= \frac{9(9t^4-6t^2+1)}{(1-3t^2)^2}=9, \tag{4} $$ hence $$ \tan^2(10)+\tan^2(50)+\tan^2(70)=9.\quad{\square} \tag{5} $$

Answer 2

Denote $\;t=20^{\LARGE\circ},\; u=\cot t,\;v=\cot 2t,\;w=\cot 4 t,\quad $ then $$S =\tan^2 10^{\LARGE\circ}+\tan^2 50^{\LARGE\circ}+\tan^2 70^{\LARGE\circ}=\cot^2 80^{\LARGE\circ}+\cot^2 40^{\LARGE\circ}+\cot^2 20^{\LARGE\circ}=u^2+v^2+w^2,$$ $$\cot8t=\cot 160^{\LARGE\circ}=-u.$$

From the known identity $$\cot(2a)=\dfrac12(\cot(a)-\tan(a))=\dfrac{\cot^2(a)-1}{2\cot(a)}\tag1$$ should $$u^2=2uv+1,\quad v^2=2vw+1,\quad w^2=-2wu+1,\tag2$$ where $$u>v>w>0.\tag 3$$ Summation of the equalities $(2)$ allows to get $$(u-v+w)^2=u^2+v^2+w^2-2v(u+w)+2uw=3,$$ $$u-v+w=\sqrt3.\tag4$$ Then $$v=\dfrac{u^2-1}{2u},\tag5$$ $$w=\sqrt3+v-u=\sqrt3+\dfrac{u^2-1}{2u}-u,$$ $$w=\sqrt3-\dfrac{u^2+1}{2u}.\tag6$$

On the other hand, $$\dfrac1{\sqrt3}=\cot 3t=\dfrac{u(u^2-3)}{3u^2-1},$$ $$u^3=\dfrac{3u^2-1}{\sqrt3}+3u\tag7.$$ Therefore, $$\begin{align} &u^2+v^2+w^2=u^2+\left(\dfrac{u^2-1}{2u}\right)^2+\left(\sqrt3-\dfrac{u^2+1}{2u}\right)^2\\[4pt] &=\dfrac1{4u^2}\left(4u^4+\left(u^2-1\right)^2+\left(2u\sqrt3-u^2-1\right)^2\right)\\[4pt] &=\dfrac1{4u^2}\left(6u^4-4u^3\sqrt3+12u^2-4u\sqrt3+2\right)\\[4pt] &=\dfrac1{4u^2}\left(6u^4-4u^3\sqrt3+12u^2-4u\sqrt3+2 -(6u+2\sqrt3)\left(u^3-\dfrac{3u^2-1}{\sqrt3}-3u\right)\right)\\[4pt] &=\dfrac{36u^2}{4u^2}=9. \end{align}$$