1.

If 4nα = π, then the value of cotα .cot2α.cot3α..............cot(2n–1)α is(a)   1(b)   –1(c)  ∞(d)  None of these

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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