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Question

For a positive integer n, let fn(θ)=(tanθ2)(1+secθ)(1+sec2θ)(1+sec22θ).....(1+sec2nθ) , then

A
f2(π16)=1
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B
f3(π32)=1
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C
f4(π64)=1
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D
f5(π128)=1
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Solution

The correct options are
A f2(π16)=1
B f3(π32)=1
C f4(π64)=1
D f5(π128)=1
1292254_1009974_ans_e235c9a016364e3eb365819f04f5ee3d.jpg

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