Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$ Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$.
Can anyone help me with this? Thank You!

Notice that for $\theta = 20, 40,$ and $80$ degrees you have $\tan^2(3\theta) = 3$. The tangent triple angle formula, which you can get from the tangent angle addition formula, says that
$$\tan(3\theta) = {3\tan(\theta) - \tan^3(\theta) \over 1 - 3 \tan^2(\theta)}$$
So the equation $\tan^2(3\theta) = 3$ can be expressed as 
$$(3\tan(\theta) - \tan^3(\theta))^2 = 3(1 - 3 \tan^2(\theta))^2$$
After a little algebra, this becomes the following, where $x = \tan(\theta)$.
$$x^6 - 33x^4 + 27x^2 - 3 = 0$$
By the above, this has roots  $x = \tan(20^\circ), \tan(40^\circ),$ and $\tan(80^\circ)$. Since $x$ only appears to even powers here, the other roots must be $x = -\tan(20^\circ), -\tan(40^\circ),$ and $-\tan(80^\circ)$.
The sum of the squares of all six roots is thus given by $2(\tan^2(20^\circ) + \tan^2(40^\circ) + \tan^2(80^\circ))$. However, if we write these roots as $r_1,...,r_6$, then we also have
$$\sum_i  r_i^2 = \left(\sum_i  r_i\right)^2 - 2\sum_{i < j} r_ir_j$$
But $\sum_i  r_i$ is the coefficient of $x^5$ in the above equation, namely zero, and 
$\sum_{i < j} r_ir_j$ is the coefficient of $x^4$, namely $-33$. So you get 
$$2(\tan^2(20^\circ) + \tan^2(40^\circ) + \tan^2(80^\circ)) = -2\times-33$$
So we conclude that
$$\tan^2(20^\circ) + \tan^2(40^\circ) + \tan^2(80^\circ) = 33$$
