1.

For a positive integer n, let fn(θ)=(tanθ2)(1+secθ)(1+sec2θ)(1+sec4θ)…(1+sec2nθ). Then

Answer»

For a positive integer n, let
fn(θ)=(tanθ2)(1+secθ)(1+sec2θ)(1+sec4θ)(1+sec2nθ).
Then




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