<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] $$\tan60^{\circ}=\sqrt3$$ gives
$$\frac{3\tan20^{\circ}-\tan^320^{\circ}}{1-3\tan^220^{\circ}}=\sqrt3,$$
which gives $$\left(\frac{3\tan20^{\circ}-\tan^320^{\circ}}{1-3\tan^220^{\circ}}\right)^2=3,$$
which gives your identity. 
</s>