`sum_(n=1)^ootan(theta/(2^n))/(2^(n-1)cos(theta/(2^(n-1))))`

Options

A. `(2)/(sin 2theta)-(1)/(theta)`
B. `(2)/(sin 2theta)+(1)/(theta)`
C. `(1)/(sin 2theta)-(1)/(theta)`
D. `(1)/(sin theta)-(1)/(theta)`

Answer

A

Explanation

`S_(N)=sum_(r=1)^(n)(tan((theta)/(2^(r))))/(2^(r-1)cos((theta)/(2^(r-1))))`
`=sum_(r=1)^(n)(sin((theta)/(2)))/(2^(r-1)cos((theta)/(2))cos((theta)/(2^(r-1))))`
`=sum_(r=1)^(n)(2sin^(2)((theta)/(2)))/(2^(r-1){2sin((theta)/(2))cos((theta)/(2))}cos((theta)/(2^(r-1))))`
`=sum_(r=1)^(n)(1-cos((theta)/(2^(r-1))))/(2^(r-1)sin((theta)/(2^(r-1)))cos((theta)/(2^(r-1))))`
`=sum_(r=1)^(n)((1)/(2^(r-1)sin((theta)/(2^(r-1))))-(1)/(2^(r-1)sin((theta)/(2^(r-1)))))`
`=(2)/(sin 2theta)-(1)/(2^(n-1)sin((theta)/(2^(n-1))))`
`therefore S_(oo)=(2)/(sin2theta)-underset(nrarroo)"lim"((theta)/(2^(n-1)).(1)/(theta))/(sin((theta)/(2^(n-1))))=(2)/(sin theta)-(1)/(theta)`