Q.

If  fn(x)=(tanx2)(1+secx)(1+sec2x)(1+sec2nx), then among the following is not true

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a

f5(π128)=0

b

f3(π32)=1

c

f2(π16)=1

d

f4(π64)=1

answer is D.

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Detailed Solution

fn(x)=(tanx2)(1+secx)(1+sec2x)(1+sec2nx)

fn(x)=sinx2cosx2(2cos2x2cosx)(2cos2xcos2x)2cos2n1xcos2nx

=2nsin(x2)cos(x2)cosxcos(2n1x)cos2nx

=tan(2nx)

f5(π128)=tan(25×π128)=tan(32π128)=1

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