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If a 1 , a 2 , a 3 are in arithmetic progression and d is the common difference, then tan - 1 ⁡ d 1 + a 1 a 2 + tan - 1 ⁡ d 1 + a 2 a 3 =

Options

  1. Atan - 1 ⁡ 2 d 1 + a 1 a 3
  2. Btan - 1 ⁡ d 1 + a 1 a 3
  3. Ctan - 1 ⁡ 2 d 1 + a 2 a 3
  4. Dtan - 1 ⁡ 2 d 1 - a 1 a 3

Correct answer

A. tan - 1 ⁡ 2 d 1 + a 1 a 3

Step-by-step solution

Since, a 1 ,   a 2 ,   a 3 are in A.P. ⇒       a 2 - a 1 = d = a 3 - a 2 Now, tan - 1 ⁡ d 1 + a 1 a 2 + tan - 1 ⁡ d 1 + a 2 a 3 = tan - 1 ⁡ a 2 - a 1 1 + a 1 a 2 + tan - 1 ⁡ a 3 - a 2 1 + a 2 a 3 ∵ tan − 1 x − tan − 1 y = tan − 1 x − y 1 + x y = tan - 1 ⁡ a 2 - tan - 1 ⁡ a 1 + tan - 1 ⁡ a 3 - tan - 1 ⁡ a 2   = tan - 1 ⁡ a 3 - tan - 1 ⁡ a 1 = tan - 1 ⁡ a 3 - a 1 1 + a 1 a 3 = tan

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