For n ∈ N, let
fₙ(θ)=tan ((θ)/(2)) (1+secθ)(1+sec 2θ)(1+sec 4θ)⋯(1+sec 2ⁿθ).
Which of the following is true?

Question

Accepted Answer

All of these. Using tan(theta/2)(1+sec theta) = tan theta repeatedly, the telescoping product gives f_n(theta) = tan(2^n theta). Then f_2(pi/16) = tan(pi/4) = 1, f_3(pi/32) = tan(pi/4) = 1, and f_4(pi/64) = tan(pi/4) = 1, so all of these are true.

For n ∈ N, let fₙ(θ)=tan ((θ)/(2)) (1+secθ)(1+sec 2θ)(1+sec 4θ)⋯(1+sec 2ⁿθ). Which of the following is true?

Solution

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