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# 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)

147 Points
11 years ago

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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Priyansh Goel
21 Points
11 years ago

Ans is ABCD.