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

Transcript

Ex 2.2, 15 If tan-1 (x β 1)/(x β 2) + tan-1 (x + 1)/(x + 2) = π/4 , then find the value ofΒ x. We know that tan-1 x + tan-1 y = tan-1 ((π± + π² )/( π β π±π²)) tan-1 ((x β 1)/(x β 2)) + tan-1 ((x + 1)/(x + 2))= tan-1 [((x β 1 )/(x β 2) + (x + 1)/(x + 2))/(1β (x β 1)/(x β 2) Γ (x + 1)/(x + 2))] = tan-1 [(((x β 1) (x + 2) + (x + 1)(x β 2))/((x β 2) (x + 2) ))/(((x β 2) (x + 2) β (x β 1) (x + 1))/((x β 2) (x + 2) ))] = tan-1 [((x β 1) (x + 2) + (x + 1)(x β 2))/((x β 2) (x + 2) ) Γ ((x β 2) (x + 2))/((x + 2) (x β 2) β (x β 1)(x + 1))] = tan-1 [((x β 1) (x + 2) + (x + 1)(x β 2))/((x + 2) (x β 2) β (x β 1)(x + 1)) ] Using (a + b) (a β b) = a2 β b2 = tan-1 [((x β 1) (x + 2) + (x + 1)(x β 2))/(π₯2 β 22 β[π₯2 β 12]) ] = tan-1 [(π₯ (x + 2) β 1 (x + 2) + x (x β 2) + 1 (x β 2))/(π₯2 β 4 βπ₯2 + 1) ] = tan-1 [(π₯2 + 2π₯ β π₯ β 2 + π₯2 β 2π₯ + π₯ β 2 )/(π₯2 β π₯2 β 4 + 1) ] Using (a + b) (a β b) = a2 β b2 = tan-1 [((x β 1) (x + 2) + (x + 1)(x β 2))/(π₯2 β 22 β[π₯2 β 12]) ] = tan-1 [(π₯ (x + 2) β 1 (x + 2) + x (x β 2) + 1 (x β 2))/(π₯2 β 4 βπ₯2 + 1) ] = tan-1 [(π₯2 + 2π₯ β π₯ β 2 + π₯2 β 2π₯ + π₯ β 2 )/(π₯2 β π₯2 β 4 + 1) ] = tan-1 [(π₯2+ π₯2+ 2π₯β 2π₯ β π₯ + π₯ β 2 β 2)/(π₯2 β π₯2 β 4 + 1) ] = tan-1 [(2x2 β4)/(β3)] β΄ tan-1 ((x β 1)/(x β 2)) + tan-1 ((x + 1)/(x + 2)) = tan-1 [(2x2 β4)/(β3)] Given tan-1 (x β 1)/(x β 2) + tan-1 (x + 1)/(x + 2) = π/4 Putting values tan-1 [(2x2 β4)/(β3)] = Ο/4 (2x2 β4)/(β3) = tan Ο/4 (2x2 β4)/(β3) = tan 45Β° (2x2 β4)/(β3) = 1 2x2 β 4 = β 3 2x2 = β 3 + 4 2x2 = 1 x2 = 1/2 x = Β± β(1/2) = Β± 1/β2 Thus, x = Β± 1/β2

Serial order wise