1.

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

Answer» 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.`


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