Correct Answer - C
We known that
`cos alpha+cos(alpha+beta)+cos(alpha+2 beta)+...+cos{alpha+(n-1)beta}`
`=(cos{alpha+(n-1)(beta)/(2)}sin((nbeta)/(2)))/(sin""(beta)/(2))`
`thereforecos""(pi)/(11)+cos""(3pi)/(11)+cos""(5pi)/(11)+cos""(7pi)/(11)+cos""(9pi)/(11)`
`=(cos((pi)/(11)+(4pi)/(11))sin((5pi)/(11)))/(sin((pi)/(11)))" "[because alpha=(pi)/(11)and beta=(2pi)/(11)]`
`=(cos""(5pi)/(11)sin""(5pi)/(11))/(sin""(pi)/(11))=1/2(sin""(10pi)/(11))/(sin""(pi)/(11))=1/2.`
