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Ex 2.2, 15 - If tan-1 (x - 1)/(x - 2) + tan-1 (x+1)/(x+2) = pi/4 - Ex 2.2

  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
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Ex 2.2, 15 If tan-1 (x βˆ’ 1)/(x βˆ’ 2) + tan-1 (x + 1)/(x + 2) = πœ‹/4 , then find the value ofΒ x. We know that tan-1 x + tan-1 y = tan-1 ((𝐱 + 𝐲 )/( 𝟏 βˆ’ 𝐱𝐲)) tan-1 ((x βˆ’ 1)/(x βˆ’ 2)) + tan-1 ((x + 1)/(x + 2))= tan-1 [((x βˆ’ 1 )/(x βˆ’ 2) + (x + 1)/(x + 2))/(1βˆ’ (x βˆ’ 1)/(x βˆ’ 2) Γ— (x + 1)/(x + 2))] = tan-1 [(((x βˆ’ 1) (x + 2) + (x + 1)(x βˆ’ 2))/((x βˆ’ 2) (x + 2) ))/(((x βˆ’ 2) (x + 2) βˆ’ (x βˆ’ 1) (x + 1))/((x βˆ’ 2) (x + 2) ))] = tan-1 [((x βˆ’ 1) (x + 2) + (x + 1)(x βˆ’ 2))/((x βˆ’ 2) (x + 2) ) Γ— ((x βˆ’ 2) (x + 2))/((x + 2) (x βˆ’ 2) βˆ’ (x βˆ’ 1)(x + 1))] = tan-1 [((x βˆ’ 1) (x + 2) + (x + 1)(x βˆ’ 2))/((x + 2) (x βˆ’ 2) βˆ’ (x βˆ’ 1)(x + 1)) ] Using (a + b) (a – b) = a2 – b2 = tan-1 [((x βˆ’ 1) (x + 2) + (x + 1)(x βˆ’ 2))/(π‘₯2 βˆ’ 22 βˆ’[π‘₯2 βˆ’ 12]) ] = tan-1 [(π‘₯ (x + 2) βˆ’ 1 (x + 2) + x (x βˆ’ 2) + 1 (x βˆ’ 2))/(π‘₯2 βˆ’ 4 βˆ’π‘₯2 + 1) ] = tan-1 [(π‘₯2 + 2π‘₯ βˆ’ π‘₯ βˆ’ 2 + π‘₯2 βˆ’ 2π‘₯ + π‘₯ βˆ’ 2 )/(π‘₯2 βˆ’ π‘₯2 βˆ’ 4 + 1) ] Using (a + b) (a – b) = a2 – b2 = tan-1 [((x βˆ’ 1) (x + 2) + (x + 1)(x βˆ’ 2))/(π‘₯2 βˆ’ 22 βˆ’[π‘₯2 βˆ’ 12]) ] = tan-1 [(π‘₯ (x + 2) βˆ’ 1 (x + 2) + x (x βˆ’ 2) + 1 (x βˆ’ 2))/(π‘₯2 βˆ’ 4 βˆ’π‘₯2 + 1) ] = tan-1 [(π‘₯2 + 2π‘₯ βˆ’ π‘₯ βˆ’ 2 + π‘₯2 βˆ’ 2π‘₯ + π‘₯ βˆ’ 2 )/(π‘₯2 βˆ’ π‘₯2 βˆ’ 4 + 1) ] = tan-1 [(π‘₯2+ π‘₯2+ 2π‘₯βˆ’ 2π‘₯ βˆ’ π‘₯ + π‘₯ βˆ’ 2 βˆ’ 2)/(π‘₯2 βˆ’ π‘₯2 βˆ’ 4 + 1) ] = tan-1 [(2x2 βˆ’4)/(βˆ’3)] ∴ tan-1 ((x βˆ’ 1)/(x βˆ’ 2)) + tan-1 ((x + 1)/(x + 2)) = tan-1 [(2x2 βˆ’4)/(βˆ’3)] Given tan-1 (x βˆ’ 1)/(x βˆ’ 2) + tan-1 (x + 1)/(x + 2) = πœ‹/4 Putting values tan-1 [(2x2 βˆ’4)/(βˆ’3)] = Ο€/4 (2x2 βˆ’4)/(βˆ’3) = tan Ο€/4 (2x2 βˆ’4)/(βˆ’3) = tan 45Β° (2x2 βˆ’4)/(βˆ’3) = 1 2x2 – 4 = βˆ’ 3 2x2 = βˆ’ 3 + 4 2x2 = 1 x2 = 1/2 x = Β± √(1/2) = Β± 1/√2 Thus, x = Β± 1/√2

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