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If 4nα = π, then the value of cotα .cot2α.cot3α..............cot(2n–1)α is(a) 1(b) –1(c) ∞(d) None of these |
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Answer» Correct option (a) 1 Explanation: cotα.cot(2n–1)α = cotα.cot(2nα – α) = cotα.cot (π/2 -α) = cotα.tanα = 1 Product of terms equidistant from the beginning and end is 1 The middle term is cotnα = cotπ/4 = 1 |
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