If `theta=pi/6` then the `10^(th)` term of the series : `1+(costheta+isintheta)^1+(costheta+isintheta)^2+........`

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

If theta=(pi)/(6) then the 10^(th) term of the series : 1+(cos theta+i sin theta)^(1)+(cos theta+i sin theta)^(2)+

if theta=(pi)/(6) then the 10 th term of the series 1+(cos theta+i sin theta)^(1)+(cos theta+i sin theta)^(2)+

If theta=(pi)/(6) then find the 10 th term of the series 1+(cos theta+i sin theta)^(1)+(cos theta+i sin theta)^(2)+...

Find (costheta+isintheta)^n

Evaluate (costheta-isintheta)^n

Evaluate (costheta-isintheta)^(-n)

Re{(1+costheta+2iSintheta)^-1}=4 then find theta

costheta/(cosec theta+1)+costheta/(cosectheta-1)=2