Answer:
Step-by-step explanation:
We have tanπ3=−tan2π3=tan4π3=3–√
Since tan(3x)=3tanx−tan3x1−3tan2x
Thus tanπ/9,−tan2π/9,tan4π/9are solutions to the polynomial 3–√=3x−x31−3x2
On simplication, x3−33–√x2−3x+3–√=0.(1)
Since (tanπ9−tan2π9+tan4π9)2
=tan2π9+tan22π9+tan24π9
+ 2( -tanπ9tan2π9+tanπ8tan4π9−tan2π9tan4π9)
By sum and product pairs of roots in (1) above
(−33–√)2=tan2π9+tan22π9+tan24π9+2×−3
∴tan2π9+tan22π9+tan24π9=33