more than one options may be correctQ. 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.