Question

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

Answer 1

Correct Answer - C
Since `tan 3theta=(3tan theta-tan^(3)theta)/(1-3tan^(2)theta)`
Putting `theta=(pi)/(9),` we get
`tan((pi)/(3))=(3tan((pi)/(9))-tan^(3)((pi)/(9)))/(1-3tan^(2)((pi)/(9)))`
or `tan^(6)((pi)/(9))-33tan^(4)((pi)/(9))+27tan^(2)((pi)/(9))=3`