For a positive integer n , let f_n(θ) =[tanθ/2](1+secθ)(1+sec2θ)(1+sec4θ)...(1+sec2ⁿθ) Then

1. f₂ [pi/16]=1
2. f₃[pi/32]=1
3. f₄[pi/64]=1
4. f₅[pi/128]=1

tanθ/2(1+1/cosθ)(1+1/cos2θ)...tanθ/2(cosθ+1)(1/cosθ)(cos2θ+1)(1/cos2θ).tanθ/2(2cos²θ/2)(1/cosθ)(2cos²θ)(1/cos2θ)=tan2θSo we can solving by taking(1+sec4θ) we will get tan4θ. So we can conclude that fn(theta)=tan(2^ntheta) so all the options are satisfying it. CORRECT ANSWER-A,B,C and D