`(5costheta+csctheta)/(5costheta-csctheta)=7/3` then `cos theta =`

costheta-cos2theta=sin3theta

Prove that, 1/(costheta-cos3theta)+1/(costheta-cos5theta)+1/(costheta-cos7theta)+…+to n terms = 1/(2sintheta) [cottheta-cot(n+1)theta]

Prove that, 1/(costheta-cos3theta)+1/(costheta-cos5theta)+1/(costheta-cos7theta)+…+to n terms = 1/(2sintheta) [cottheta-cot(n+1)theta]

Find the sum of n terms the series: 1/(costheta+cos3theta)+1/(costheta+cos5theta)+1/(costheta+cos7theta)+…

Find the sum of n terms the series: 1/(costheta+cos3theta)+1/(costheta+cos5theta)+1/(costheta+cos7theta)+…

(cos3theta+isin3theta)^5/(costheta+isintheta)^6

Prove that: (cos6theta+6cos4theta+15 cos2theta+10)/(cos5theta+5cos3theta+10 costheta)=2costheta

Solve the equation z^8+1=0 and deduce that cos4theta=8(costheta-cos(pi/8))(costheta-cos((3theta)/3))(costheta- cos((5pi)/8))(costheta-cot((7pi)/8))

If A = [(sintheta ,- costheta), (costheta, sintheta)] and B = [(costheta , sin theta),(-sintheta, cos theta)] . Compute (sin theta) A + (cos theta) B.

(csctheta-sintheta")(sectheta-costheta)(tantheta+cottheta)=