<s>[INST] Prove that: $\tan^6 20° - 33\tan^4 20° + 27\tan^2 20°=3$ Prove that: $\tan^6 20° - 33\tan^4 20° + 27\tan^2 20°=3$
My Attempt:
$$L.H.S=\tan^6 20 - 33\tan^4 20 + 27\tan^2 20°$$
$$=\tan^2 20°(\tan^4 20° - 33\tan^2 20°+27)$$
$$=(\sec^2 20° -1)(\tan^4 20° - 33\tan^2 20° + 27)$$
Please help me to continue from here..
 [/INST] My idea: We know formulas for $\cos(20°)$; we can write it as a relation for $\sec(20°)$, and then use the identity $\sec^2(x)=1+\tan^2(x)$  

First of all, we'll compute some formulas related to $\cos(20°)$ (well-known, but I don't renember them xD). We know that
$$ \cos(3x)=\cos(2x)\cos(x)-\sin(x)\sin(2x)=(2\cos^2(x)-1)\cos(x)-2\sin^2(x)\cos(x)$$
So, $\cos(3x)=2\cos^3(x)-\cos(x)-2(1-\cos^2(x))\cos(x)=4\cos^3(x)-3\cos(x)$. Plugging $x=20°$, and calling $a=\cos(20)$, we get
$\frac{1}{2}=\cos(60°)=4\cos^3(20^°)-3\cos(x)=4a^3-3a$
So, $1=8a^3-6a$. Squaring both sides, we get $1=64a^6-96a^4+36a^2$. Now, dividing by $a^6$ both sides, and calling $b=\frac{1}{a}=\frac{1}{\cos(20^°)}=\sec(20°)$, we have
$$ \frac{1}{a^6}=64-\frac{96}{a^2}+\frac{36}{a^4} \Rightarrow b^6=64-96b^2+36b^4  $$
The key step now is try to use the identity $\sec^2(x)=1+\tan^2(x)$. Calling $c=\tan(20^°)$, we can plug $b^2=1+c^2$ on our formula, to get
$$ \begin{align*}
(1+c^2)^3 &= 64-96(1+c^2)+36(1+c^2)^2 \\
1+3c^2+3c^4+c^6 &= 64-96(1+c^2)+36(1+2c^2+c^4) \\
1+3c^3+3c^4+c^6 &= 64-96-96c^2+36+72c^2+36c^4 \\
1+3c^3+3c^4+c^6 &= 4-24c^2+36c^4 \\
c^6-33c^4+27c^2 &= 3 \\
\tan(20°)^6-33\tan(20°)^4+27\tan(20°)^2 &= 3 \\
\end{align*}$$
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