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JEE Main202227 Jun 2022Evening ShiftMathematicsInverse Trigonometric FunctionsActual

The value of cot ∑ n = 1 50 tan - 1 1 1 + n + n 2 is

Options

  1. A25 26
  2. B50 51
  3. C26 25
  4. D52 51

Correct answer

C. 26 25

Step-by-step solution

Now ∑ n = 1 50 tan - 1 1 1 + n + n 2 = ∑ n = 1 50 tan - 1 n + 1 - n 1 + n n + 1 = ∑ n = 1 50 tan - 1 n + 1 - tan - 1 n = tan - 1 51 - tan - 1 1 = tan - 1 51 - 1 1 + 51 = tan - 1 50 52 = cot - 1 26 25 So cot ∑ n = 1 50 tan - 1 1 1 + n + n 2 = cot cot - 1 26 25 = 26 25

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