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Example 7 - Show that tan-1 x + tan-1 2x/(1-x2) - Inverse

Example 7 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2

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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) Solving 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‘₯(1 βˆ’ π‘₯2) + 2π‘₯)/(1 βˆ’ π‘₯2))/(γ€–(1 βˆ’ π‘₯2) βˆ’ 2π‘₯γ€—^2/(1 βˆ’ π‘₯2)) We know that tan-1 x + tan-1 y = tan-1 ((𝒙+π’š )/(𝟏 βˆ’π’™π’š)) Replacing x by x and y by 2π‘₯/(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 ((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 Thus L.H.S = R.H.S Hence proved

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