<s>[INST] Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$ Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$.
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 [/INST] 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 $\boxed{\tan ^ { 2 } 20 ^ { \circ } + \tan ^ { 2 } 4 0 ^ { \circ } + \tan ^ { 2 } 80 ^ { \circ } = 33.}$
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