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cos^6pi/9-33tan^4pi/9+27tan^2pi/9 is equ...

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

A

0

B

`sqrt(3)`

C

3

D

9

Text Solution

Verified by Experts

The correct Answer is:
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`
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