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

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