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..

Let $\pm\sqrt3=\pm\tan60^\circ=\tan3x=\dfrac{3t-t^3}{1-3t^2}$ where $t=\tan x$
$$\implies t^3-3t=\pm\sqrt3(1-3t^2)$$
Square both sides
OR observe that $$\dfrac{t^6-33t^4+27t^2-3}{t^3-3t\mp\sqrt3(1-3t^2)}=t^3-3t\pm\sqrt3(1-3t^2)$$
So, $3x=180^\circ n\pm60^\circ$ where $n$ is any integer.
$\implies x=60^\circ n\pm20^\circ$
Here $n=0$
