If `(costheta+isintheta)(cos2theta+isin2theta)…..(cosntheta+isinntheta)=1`, then the value of `theta`, is

Options

A. `4mpi`
B. `(2mpi)/(n(n+1))`
C. `(4mpi)/(n(n+1))`
D. `(mpi)/(n(n+1))`

Answer

C

Explanation

We have,
`(costheta+isintheta)(cos2theta+isin2theta)…..(cosntheta+isinntheta)=1`
`rArr cos(theta+2theta+3theta+……+ntheta)+isin(theta+2theta+……..+ntheta)=1`
`rArr cos{(n(n+1))/(2)theta}=1` and, `sin{(n(n+1)theta)/(2)}=0`
`rArr (n(n+1))/(2)=theta=2mpi rArr theta=(4mpi)/(n(n+1))`, where `m in Z`.