Example 7 - Show that tan-1 x + tan-1 2x/(1-x2) - Inverse - Not clear how to approach

  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
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Example 7 Show that tan-1 π‘₯ + tan-1 2π‘₯/(1 βˆ’π‘₯2) = tan-1 (3π‘₯ βˆ’ π‘₯3)/(1 βˆ’ 3π‘₯2) Taking L.H.S tan-1 π‘₯ + tan-1 2π‘₯/(1 βˆ’ π‘₯2) = tan-1 (π‘₯ + 2π‘₯/(1 βˆ’ π‘₯2))/(1βˆ’ π‘₯ Γ— 2π‘₯/(1 βˆ’ π‘₯2)) = tan-1 (π‘₯ + 2π‘₯/(1 βˆ’ π‘₯2))/(1βˆ’ γ€–2π‘₯γ€—^2/(1 βˆ’ π‘₯2)) = tan-1 ((π‘₯(1 βˆ’ π‘₯2) + 2π‘₯)/(1 βˆ’ π‘₯2))/(γ€–(1 βˆ’ π‘₯2) βˆ’ 2π‘₯γ€—^2/(1 βˆ’ π‘₯2)) = tan-1 ((π‘₯ βˆ’ π‘₯3 + 2π‘₯)/(1 βˆ’ π‘₯2))/(γ€–1 βˆ’ π‘₯2βˆ’ 2π‘₯γ€—^2/(1 βˆ’ π‘₯2)) = tan-1 ((π‘₯ + 2π‘₯ βˆ’ π‘₯3)/(1 βˆ’ π‘₯2))/(γ€–1 βˆ’ π‘₯2βˆ’ 2π‘₯γ€—^2/(1 βˆ’ π‘₯2)) = tan-1 ((3π‘₯ βˆ’ π‘₯3)/(1 βˆ’ π‘₯2))/(γ€–1 βˆ’ 3π‘₯γ€—^2/(1 βˆ’ π‘₯2)) = tan-1 (3π‘₯ βˆ’ π‘₯3)/(1 βˆ’ π‘₯2) Γ— (1 βˆ’ π‘₯2)/γ€–1 βˆ’ 3π‘₯γ€—^2 = tan-1 (3π‘₯ βˆ’ π‘₯3)/γ€–1 βˆ’ 3π‘₯γ€—^2 = R.H.S Hence, L.H.S = R.H.S Hence proved

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