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Ex 2.2, 4 - Prove 2tan-1 1/2 + tan-1 1/7 = tan-1 31/17 - Formulae based

  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
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Ex 2.2, 4 Prove 2tan-1 1/2 + tan-1 1/7 = tan-1 31/17 First we calculate value of 2tan-1 𝟏/𝟐 We know that 2tan-1x = tan-1 ((𝟐𝐱 )/( 𝟏 βˆ’ 𝐱^𝟐 )) Replacing x with 1/2 2tan-1 1/2 = tan-1 (2 Γ— 1/2)/(1 βˆ’ (1/2)2) = tan-1 (1/(1 βˆ’ 1/4)) = tan-1 (1/((4 βˆ’ 1)/4)) = tan-1 (1/(3/4)) = tan-1 (4/3) Taking L.H.S. 2tan-1 1/2 + tan-1 1/7 Putting value of 2tan-1 1/2 = tan-1 4/3 + tan-1 1/7 = tan-1 ((4/3 + 1/7 )/( 1βˆ’ 4/3 Γ— 1/7)) = tan-1 (((4 Γ— 7 +3 Γ— 1 )/( 7 Γ— 3) )/( (7 Γ— 3 βˆ’ 4)/(7 Γ— 3))) = tan-1 (((28 + 3 )/( 21) )/( ( 21 βˆ’ 4)/21)) = tan-1 ((31/( 21) )/(17/21)) = tan-1 (31/21 Γ— 21/17)= tan-1 (31/17) = R.H.S. Hence, L.H.S. = R.H.S. Hence Proved

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