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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

Mohammed Tameem Mohiuddin , 9 Years ago
Grade 11
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