fn(θ)=∑k=1ntanθ2ksecθ2k-1, then which of the following is/are correct?

tan⁡α2cos⁡α=sin⁡α−α2cos⁡α2cos⁡α=tan⁡α−tan⁡α2now fnθ=∑k=1n tanθ2k-1-tanθ2k=tanθ20-tanθ21+tanθ21-tanθ22+tanθ22-tanθ23+........+tanθ2n-1-tanθ2n∴fn(θ)=tan⁡θ−tan⁡θ2noption 1:  f32π=tan2π-tan2π8=-1option2 :  f44π3=tan4π3-tan4π316=3-2-3=23-2 , similarly remaining two options.