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Question

If (1cot5)(1cot10)(1cot15)(1cot40)=2k, then the value of k is

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Solution

(1cotθ)(1cot(45θ))=(1cotθ)(1(cotθ+1cotθ1))=2

So,
(1cot5)(1cot10)(1cot15)...(1cot40)
=(1cot5)(1cot40)(1cot10)(1cot35) (1cot15)(1cot30)(1cot20)(1cot25)=24k=4

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