In a `Delta ABC`, if `cosA cos B cos C= (sqrt3-1)/(8) and sin A sin B sin C= (3+ sqrt3)/(8)`, then The value of `tan A tan B tanC ` is

Options

A. `5-4sqrt3`
B. `5+4sqrt3`
C. `6+sqrt3`
D. `6-sqrt3`

Answer

B

Explanation

`because cos A cos B cos C = (sqrt3 -1)/(8)`
`sin A sin B sin C = (3+sqrt3)/(8)`
`therefore tanA tan B tan C = (3+ sqrt3)/(sqrt3-1)" " `...(1)
`therefore tan A+ tanB + tan C = tan A tan B tan C `
`" " = (3 + sqrt3)/(sqrt3 -1) " "` ...(2)
Now `A + B +C= pi`
`therefore cos(A+B +C) = -1 `
`therefore cos A cos B cos C [1 -sum tan A tan B] =-1`
`therefore sum tanA tan B = 5 + 4sqrt3" "` ...(3)
Form (1), (2) and (3), we get
`tanA, tan B, tanC` are roots of
`x^(3)- ((3+sqrt3)/( sqrt3-1)) x^(2) + (5+4sqrt3)x - ((3+sqrt3))/(sqrt3-1)=0`
or `x^(3) - (2+ sqrt3)sqrt3 x^(2) + (5+4sqrt3)x - (2+ sqrt3) sqrt3 =0`
or ` x^(3) - (3 + 2 sqrt3)x^(2) + (5+ 4sqrt3)x - (3+ 2sqrt3)=0`
or `(x-1)(x-sqrt3)(x-(2+sqrt3)) =0`
`therefore tanA =1, tan B = sqrt3, tan C = 2 + sqrt3`