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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



A

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B

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C

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D

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Solution

The correct options are
A


B


C


D


(a,b,c,d) fn(θ)=sin(θ2)cos(θ2)[2 cos2 θ2cos θ.2 cos2 θcos 2θ.2 cos2 θcos 4θ..]

Combine first two factors, fn(θ)=sin θcos θ[2 cos2 θcos 2θ32 cos2 2θcos 4θ...]

Again combine first two factors,

fn(θ)=tan 2θ[2 cos2 2θcos 4θ...]=tan(2nθ) f2(π16)=tan4π16=tan(π4)=1f3(π32)=tan8π32=tan(π4)=1

f4(π64)=tan16π64=tan(π4)=1f5(π128)=tan 32π128=tan(π4)=1


Mathematics

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