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For a positive integer n, let fn(θ)=(tanθ2)(1+sec θ)(1+sec 2θ)(1+sec 4θ)......(1+sec 2nθ). Then |
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Answer» For a positive integer n, let fn(θ)=(tanθ2)(1+sec θ)(1+sec 2θ)(1+sec 4θ)......(1+sec 2nθ). Then |
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