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Ex 2.2
Ex 2.2, 2 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 3 Deleted for CBSE Board 2022 Exams
Ex 2.2, 4 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 5 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 6 Deleted for CBSE Board 2022 Exams
Ex 2.2, 7 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 8 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 9 Deleted for CBSE Board 2022 Exams
Ex 2.2, 10 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 11 Deleted for CBSE Board 2022 Exams
Ex 2.2, 12 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 13 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 14 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 15 Important Deleted for CBSE Board 2022 Exams You are here
Ex 2.2, 16 Deleted for CBSE Board 2022 Exams
Ex 2.2, 17 Deleted for CBSE Board 2022 Exams
Ex 2.2, 18 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 20 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 21 (MCQ) Deleted for CBSE Board 2022 Exams
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