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!
 A: 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
$$\tan^2(20^\circ) + \tan^2(40^\circ) + \tan^2(80^\circ) = 33$$
A: Here's a linear algebraic route: from this answer, we find that the eigenvalues of the $4\times4$ min-matrix
$$\mathbf M=\begin{pmatrix}
 1 & 1 & 1 & 1 \\
 1 & 2 & 2 & 2 \\
 1 & 2 & 3 & 3 \\
 1 & 2 & 3 & 4
\end{pmatrix}$$
are $\lambda_k=\dfrac14\sec^2\left(\dfrac{k\pi}{9}\right)$ for $k=1,\dots,4$. From this, we have that the eigenvalues of $4\mathbf M-\mathbf I$ are $\nu_k=\tan^2\left(\dfrac{k\pi}{9}\right)$, and since the sum of the eigenvalues is equal to the trace of the matrix,
$$\tan^2\frac{\pi}{9}+\tan^2\frac{2\pi}{9}+\tan^2\frac{4\pi}{9}=4(1+2+3+4)-4-\tan^2\frac{\pi}{3}=33$$
A: In $(7)$ from this answer, it is shown that
$$
\sum_{l=1}^n\tan^2\left(\frac{\pi l}{2n+1}\right)=n(2n+1)
$$
In this case, $n=4$ and you're missing $l=3$. $\tan^2(60^\circ)=3$, so the sum would be
$$
36-3=33
$$
A: Since $$1+\tan^2 \alpha=\frac 1{\cos^2\alpha}=\frac 2{1+\cos 2\alpha},$$
$$\tan ^{2}20^{\circ}+\tan ^{2}40^{\circ}+\tan ^{2}80^{\circ}+3=
\frac 2{1+\cos 40^{\circ}}+\frac 2{1+\cos 80^{\circ}}+\frac 2{1+\cos 160^{\circ}}.$$
Reducing the last sum to a common denominator and using the equalities from the appendix of the answer by lab bhattacharjee, we obtain $36$.
A: Using the formula $ \cos 3 \theta = 4 \cos ^ { 3 } \theta - 3\cos \theta,$ we have
$\qquad\qquad 4 \cos ^ { 3 } 40 ^ { \circ } - 3 \cos 40 ^ { \circ }$
$\left. { = \cos 120 ^ { \circ } } { = - \frac { 1 } { 2 } } \right. $
$\qquad \Rightarrow \quad 8 \cos ^ { 3 } 40 ^ { \circ } - 6 \cos 40 ^ { \circ } + 1 = 0 $$ \quad \Rightarrow \quad  \cos 40 ^ { \circ } $ is a root of $8 x ^ { 3 } - 6 x + 1 = 0 \cdots (1) $
$\text{Similarly, }$$\cos 80 ^ { \circ }\text{and}\cos 160^{ \circ }\text{are also the roots of (1).}$$\text{Now we are going to transform (1) by $ y=x^2  $ into another equation (2) whose roots are } 
$$\cos ^2 40 ^ { \circ }, \cos ^2 80 ^ { \circ } \text{ and } \cos ^2160^ { \circ }\tag*{}.  $
$\text{From (1),} -1= x(8x^2–6). \text{ Squaring both sides, we have }$$1=x^2(8x^2–6)^2 
\text{and hence }64y^3–96y^2+36y-1=0 \tag*{(2)} \\$$\text{whose roots are  }\cos ^2 40 ^ { \circ }, \cos ^2 80 ^ { \circ } \text{and }\cos ^2 20^ { \circ }.  $
However, in order to get the sum of squares of tangents of $20^{\circ}, 40^{\circ} \text{and }80^{\circ},$ we need their corresponding secant squares.
One more transformation $z=\frac{1}{y}$ is introduced to get another equation (3):
$$\quad z ^ { 3 } - 36 z^2 + 96 z - 64 = 0, $$
$\text{whose roots are }\sec ^ { 2 } 20 ^ { \circ } , \sec ^ { 2 }40 ^ { \circ } \textrm{and } \sec ^ { 2 } 80 ^ { \circ }.$
$\therefore \tan ^ { 2 } 20 ^ { \circ } + \tan ^ { 2 } 40 ^ { \circ } + \tan ^ { 2 } 80 ^ { \circ } + 3$
$=\tan ^ { 2 } 20^ { \circ } +1 + \tan ^ { 2 } 40 ^ { \circ } +1 + \tan ^ { 2 } 80 ^ { \circ } + 1$
$= \sec ^ { 2 } 20 ^ { \circ } + \sec ^ { 2 } 40 ^ { \circ } + \sec ^ { 2 } 80 ^ { \circ }$
$=36$
Hence $\boxed{\tan ^ { 2 } 20 ^ { \circ } + \tan ^ { 2 } 4 0 ^ { \circ } + \tan ^ { 2 } 80 ^ { \circ } = 33.}$
