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JEE Advanced2013MathematicsInverse Trigonometric FunctionsActual

The value of c o t Σ n = 1 2 3 cot - 1 1 + Σ k = 1 n 2 k is

Options

  1. A2 3 2 5
  2. B2 5 2 3
  3. C2 3 2 4
  4. D2 4 2 3

Correct answer

B. 2 5 2 3

Step-by-step solution

c o t   Σ n = 1 23 cot - 1 1 + Σ k = 1 n 2 k = cot Σ n = 1 2 3 cot - 1 1 + 2 + 4 + 6 + … + 2 n =cot Σ cot - 1 1 + n n + 1 =cot Σ tan - 1 n + 1 - n 1 + n n + 1 c o t - 1 x = tan - 1 1 x =cot Σ n = 1 2 3 tan - 1 n + 1 - tan - 1 n tan -1 x-y 1 + x y =tan -1 x-tan -1 y Now tan - 1 n + 1 - tan - 1 n=tan - 1 2 - tan - 1 1+tan - 1 3 - tan - 1 2+....+tan - 1 24 - tan - 1 23 =cot tan - 1 2 4 - tan - 1 1 =cot tan - 1 2 4 - 1 1 + 2 4 =cot cot - 1 2 5 2 3 = 2 5 2 3

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