Prove that `(4cos^2 9^0-3)(4cos^2 27^0-3)=tan9^0dot`

Explanation

`L.H.S. = (4cos^2 9^@ - 3)(4cos2 27^@ - 3)`

`=(cos9^@ (4cos^2 9^@ - 3))/(cos9^@) *(cos27^@ (4cos^2 27^@ - 3))/cos27^@`

`=((4cos^3 9^@ - 3cos9^@)(4cos^3 27^@ - 3cos27^@))/(cos9^@cos27^@)`

As `4cos^3theta - 3cos theta = cos3 theta`, so it becomes,

`=(cos27^@cos81^@)/(cos9^@cos27^@)`

`=(cos81^@)/(cos9^@)`

`=(cos(90-9)^@)/(cos9^@)`

`=sin9^@/cos9^@`

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