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Let α = √3 sin(π/9) [ 4 + sec (π/9) ]

α  × cos (π/9) = 4√3 sin(π/9) cos (π/9)  + √3 sin(π/9)

α  × cos (π/9) = 2√3 sin(2π/9)   + √3 sin(π/9)

α  × cos (π/9) - √3 sin(π/9) =  2√3 sin(2π/9)  

[ α / (2√3 ) ] × cos (π/9) - (1/2) sin(π/9) =   sin(2π/9)  .................. (1)

Let  sinβ = [ α / (2√3 ) ]  and  cosβ = (1/2)  so that tanβ = α / √3 .

By using sinβ and cosβ , we rewrite eqn.(1) as

sinβ cos(π/9) - cosβ sin(π/9) = sin ( 2π/9 )

sin[ β - ( π/9 ) ] = sin( 2π/9 )

from above expression , we get

[ β - ( π/9 ) ] = ( 2π/9 )

 β  =  π/3

since tanβ = α / √3  , we get

( α / √3 ) = tan ( π/3 ) = √3

Hence , α = 3

Answer :-

√3 sin(π/9) [ 4 + sec (π/9) ] = 3