more than one options may be correct

Q. for an +ve integer n, let _{fn(x)= {tan(x/2)}(1+secx)(1+sec2x)(1+sec4x).............(1+sec2ⁿx). Then}

(a)f₂(pie/16)

(b)f3(pie/32)

(c)f4(pie/64)

(d)f5(pie/128)

Dear raman

fn(x)= {tan(x/2)}(1+secx)(1+sec2x)(1+sec4x).............(1+sec2ⁿx).

simplyfy first bracket 

       =tanx/2 (2cos²(x/2) /cosx))(1+sec2x)(1+sec4x).............(1+sec2ⁿx).

        =tanx (1+sec2x)(1+sec4x).............(1+sec2ⁿx).

 similery solve other bracket also 

  fn(x) =tan2ⁿx  

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Ans is ABCD.