Prove that `sum_(r=1)^n(1/(costheta+"cos"(2r+1)theta))=(sinntheta)/(2sintheta costhetacos (n+1)theta),(w h e r en in N)dot`

Explanation

`S=sum_(r=1)^(n)((1)/(cos theta+cos(2r+1)theta))`
`=sum_(r=1)^(n)((sin theta)/(2cos(r+1)thetacosrthetasin theta))`
`=(1)/(2sin theta)(sum_(r=1)^(n)(sin(r+1)theta-rtheta)/(cos(4+1)theta cos rtheta))`
`=(1)/(2sin theta)(sum_(r=1)^(n)(sin(r+1)theta cos rtheta-sinrthetacos(r+1)theta)/(cos (r+1)thetacos rtheta)`
`=(1)/(2sin theta)(sum_(r=1)^(n)(r+1)theta-tan rtheta)`
`=(1)/(2sin theta)(tan (n+1)theta-tan theta)`
`=(sin n theta)/(2 sin theta. cos theta cos (n+1)theta)`