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1.For a positive integer n,let f n (θ)=(tan θ/2)(1+secθ)(1+sec 1 θ)(1+sec 2 θ)…(1+sec2 n θ) then (A)f 2 (Π/16)=1 (B)f 3 (Π/32)=1 (C)f 4 (Π/64)=1 (D)f 5 (Π/128)=1

1.For a positive integer n,let fn(θ)=(tan θ/2)(1+secθ)(1+sec1θ)(1+sec2θ)…(1+sec2nθ) then

(A)f2(Π/16)=1     (B)f3(Π/32)=1     (C)f4(Π/64)=1    (D)f5(Π/128)=1

Grade:11

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