`1/(cosalpha+cos3alpha)+1/(cosalpha+cos5alpha)....1/(cosalpha+cos(2n+1)alpha`

(1)/(cos alpha+cos3 alpha)+(1)/(cos alpha+cos5 alpha)...(.1)/(cos alpha+cos(2n+1)alpha)

1/(cos alpha + cos 3 alpha)+ 1/(cos alpha + cos 5 alpha)+ 1/(cos alpha + cos 7 alpha) +....+ 1/(cos alpha +cos(2n+1) alpha) = 1/2 "cosec" alpha[tan (n+1) alpha - tan alpha ]

Let f_n(a)=(sinalpha+sin3alpha+sin5alpha+...+sin(2n-1)alpha)/(cosalpha+cos3alpha+cos5alpha+...+cos(2n-1)alpha) Then, the value of f_4(pi/32) is equal to

Let f_n(a)=(sinalpha+sin3alpha+sin5alpha+...+sin(2n-1)alpha)/(cosalpha+cos3alpha+cos5alpha+...+cos(2n-1)alpha) Then, the value of f_4(pi/32) is equal to

Prove the following : (cos7alpha+cos3alpha-cos5alpha-cosalpha)/(sin7alpha-sin3alpha-sin5alpha+sinalpha) = cot2alpha

Prove that cosalpha/(sin3alpha)+(cos3alpha)/(sin9alpha)+(cos9alpha)/(sin27alpha)=1/2[cotalpha-cot27alpha]

prove that cosalpha\ cos2alpha\ cos4alpha......cos(2^(n-1)alpha)=(sin2^nalpha)/(2^n sinalpha)\ for\ a l l\ n in N

If tantheta = (sinalpha-cosalpha)/(sinalpha+cosalpha), then (A) sin alpha-cos alpha=+-sqrt(2) sin theta (B) sinalpha+cosalpha=+-sqrt(2) cos theta (C) cos2theta=sin2alpha (D) sin2theta+cos2alpha=0