If `alpha=[2pi]/7` then `tanalphatan2alpha+tan2alphatan4alpha+tan2alphatan4alpha=`

Options

A. `-5`
B. `-3`
C. `-1`
D. `-7`

Answer

D

Explanation

Let `A=alpha, B=2alpha, C=4alpha`
`therefore A+B+C=7alpha=2pi`
`therefore cos (A+B+C)=cos 2pi=1`
`cos A.cos B.cos C-sin sin A.sin B. cos C - sin B. sin C.cos A-sin sin C.sin A.cos B=1`
Dividing throughout by cos A.cos B. cos C, we get
`1-(tan A tan B + tan C + tan C tan A)`
`=(1)/(cos A.cos B.cos C)`
`therefore tan A tan B + tan B tan C+ tan C tan A`
`=1-(1)/(cos alpha.cos 2alpha.cos 4 alpha)`
`=1-(1)/((sin 2^(3)alpha)/(2^(3)sin alpha))`
`=1-(8 sin alpha)/(sin 8 alpha)=1-(8sin.(2pi)/(7))/(sin.(16pi)/(7))`
`= 1-8=-7`
`[because sin.(16pi)/(7)=sin(2pi + (2pi)/(7))=sin.(2pi)/(7)]`