If 4nα=π then cotα⋅cot2α⋅cot3α⋅...cot(2n−1)α is equal to
A
1
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B
−1
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C
∞
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D
none of these
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Solution
The correct option is A 1 cotα.cot2α.cot3α...cot(2n−1)α=cosα.cos2α.cos3α...cos(2n−1)αsinα.sin2α.sin3α...sin(2n−1)α=cosπ4ncos2π4ncos3π4n...cos(2n−1)π4nsinπ4nsin2π4nsin3π4n...sin(2n−1)π4n=cosπ4ncos2π4ncos3π4n...cosnπ4n...sin3π4nsin2π4nsinπ4nsinπ4nsin2π4nsin3π4n...sinnπ4n...cos3π4ncos2π4ncosπ4n=1