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Question
For a positive integer n, let
  fn(θ) = (tanθ2)(1 + sec θ)(1 + sec 2θ)(1 + sec 4θ) .........
(1 + sec 2nθ). Then
Options:
A .  f2(π16) = 1
B .  f3(π32) = 1
C .  f4(π64) = 1
D .  All of above
Answer: Option D
:
D
(a,b,c) fn(θ) = sin(θ/2)cos(θ/2)[2cos2θ/2cosθ.2cos2θcos2θ.2cos22θcos4θ]
Combine first two factors,fn(θ) = sinθcosθ[2cos2θcos2θ.2cos22θcos4θ]
Again combine first two factors,
fn(θ) = tan 2θ[2cos22θcos4θ.......] = tan (2nθ)
f2(π16) = tan 4π16 = tan (π4)= 1
f3(π32) = tan 8π32 = tan (π4) = 1
f4(π64) = tan 16π64 = tan (π4) = 1
f5(π128) = tan 32π128 = tan (π4) = 1

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