Ex 2.2

Ex 2.2,1
Deleted for CBSE Board 2023 Exams

Ex 2.2, 2 Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 3 Deleted for CBSE Board 2023 Exams

Ex 2.2, 4 Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 5 Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 6 Deleted for CBSE Board 2023 Exams

Ex 2.2, 7 Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 8 Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 9 Deleted for CBSE Board 2023 Exams

Ex 2.2, 10 Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 11 Deleted for CBSE Board 2023 Exams

Ex 2.2, 12 Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 13 Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 14 Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 15 Important Deleted for CBSE Board 2023 Exams You are here

Ex 2.2, 16 Deleted for CBSE Board 2023 Exams

Ex 2.2, 17 Deleted for CBSE Board 2023 Exams

Ex 2.2, 18 Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 19 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 20 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 21 (MCQ) Deleted for CBSE Board 2023 Exams

Chapter 2 Class 12 Inverse Trigonometric Functions

Serial order wise

Last updated at May 12, 2021 by Teachoo

Ex 2.2, 15 If tanβ1 (x β 1)/(x β 2) + tanβ1 (x + 1)/(x + 2) = π/4 , then find the value of x. Given tanβ1 ((π± β π)/(π± β π)) + tanβ1 ((π± + π)/(π± + π)) = π/4We know that tanβ1 x + tanβ1 y = tanβ1 ((π± + π² )/( π β π±π²)) Replacing x by (π₯ β 1)/(π₯ β 2) and y by ((π₯ + 1)/(π₯ + 2)) tanβ1 [((x β 1 )/(x β 2) + (x + 1)/(x + 2))/(1β (x β 1)/(x β 2) Γ (x + 1)/(x + 2))]=" " π/4 tanβ1 [((x β 1 )/(x β 2) + (x + 1)/(x + 2))/(1β (x β 1)/(x β 2) Γ (x + 1)/(x + 2))]="tan " π/4 = 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) ))]((x β 1 )/(x β 2) + (x + 1)/(x + 2))/(1β (x β 1)/(x β 2) Γ (x + 1)/(x + 2)) = "tan " π /π (((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) )) = 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)) = 1 ((x β 1) (x + 2) + (x + 1)(x β 2))/((x + 2) (x β 2) β (x β 1)(x + 1)) = 1 Using (a + b) (a β b) = a2 β b2 ((x β 1) (x + 2) + (x + 1)(x β 2))/(π₯2 β 22 β[π₯2 β 12]) = 1 (π₯ (π₯ + 2) β 1 (π₯ + 2) + π₯ (π₯ β 2) + 1 (π₯ β 2))/(π₯2 β 4 β π₯2 + 1) = 1 (π₯2 + 2π₯ β π₯ β 2 + π₯2 β 2π₯ + π₯ β 2 )/(π₯2 β π₯2 β 4 + 1) = 1 (2x2 β4)/(β3) = 1 2x2 β 4 = β3 2x2 = β3 + 4 2x2 = 1 x2 = 1/2 β΄ x = Β± π/βπ