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Question

If 4nα=π, then the value of cotαcot2αcot3α....cot(2n1)α is

A
0
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B
1
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C
2
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D
None of these
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Solution

The correct option is B 1
we can write n=π4α

Now, cot(2n1)α=cot(2π4α1)α=cot(π2α)=tanα

Similarly, cot(2n2)α=cot(2π4α2)α=cot(π22α)=tan2α and so on

So, cotα.cot2α..............cot(2n2)α.cot(2n1)α=cotα.cot2α...........tan2αtanα=(cotαtanα)(cot2αtan2α)(cot3αtan3α).......=1.1.1...............=1

Therefore, Correct Answer is 1

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