Question

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 1

Correct option is (a) 1

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 = cot π/4 = 1

Given expression = 1.1.1......................n times = 1