`tan^6(pi/9)-33tan^4(pi/9)+27tan^2(pi/9)` is equal to (a)0 (b) `sqrt3` (c) 3(d)9

Explanation

We know, `tan 3theta = (3tantheta - tan^3theta)/(1-3tan^2theta)`

If `theta = pi/9`

`=>tan(pi/3) = (3tan(pi/9) - tan^3(pi/9))/(1-3tan^2(pi/9))`

`=>(sqrt3(1-3tan^2(pi/9))) = (3tan(pi/9) - tan^3(pi/9))`

Squaring both sides,

`=>3(1+9tan^4(pi/9) - 6tan^2(pi/9)) = 9tan^2(pi/9)+tan^6pi/9 - 6tan^4(pi/9)`

`=>3+33tan^4(pi/9)-27tan^2(pi/9) = tan^6(pi/9)`

`=>tan^6(pi/9)-33tan^4(pi/9)+27tan^2(pi/9) = 3.`