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For a positive integer n , let fn(θ) =[tanθ/2](1+secθ)(1+sec2θ)(1+sec4θ)...(1+sec2nθ) Then f2 [pi/16]=1 f3[pi/32]=1f4[pi/64]=1f5[pi/128]=1

For a positive integer n , let fn(θ) =[tanθ/2](1+secθ)(1+sec2θ)(1+sec4θ)...(1+sec2nθ) Then f2 [pi/16]=1 f3[pi/32]=1f4[pi/64]=1f5[pi/128]=1

Grade:11

1 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
2 years ago
Dear student

Please complete the question

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