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  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise

Transcript

Example 13 Solve tan–1 2x + tan–1 3x = π/4 We know that tan–1 x + tan–1 y = tan–1 ((𝐱 + 𝐲)/(𝟏 − 𝐱𝐲)) tan–1 2x + tan 3x = tan–1 ((2x + 3x)/(1 − 2x × 3x)) = tan–1 (5x/(1 − 6x2)) Now, given that tan–1 2x + tan–1 3x = π/4 tan–1 (5x/(1 − 6x2)) = π/4 5x/(1 − 6x2) = tan π/4 5x/(1− 6x2) = 1 5x = 1 × (1 – 6x2) 5x = 1 – 6x2 6x2 + 5x – 1 = 0 6x2 + 6x – x – 1 = 0 6x(x + 1 ) – 1 (x – 1) = 0 (6x – 1) (x + 1) = 0 Thus, x = 1/6 or x = −1 But for x = −1 tan–1 2x + tan–1 3x = π/4 tan–1 (–2) + tan–1 (–3) = π/4 So, L.H.S becomes negative but R.H.S is positive. So, x = –1 is not possible. Hence, x = 𝟏/𝟔 is the only solution of the given equation

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.