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Ex 2.2

Ex 2.2,1
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Ex 2.2, 15 Important Deleted for CBSE Board 2022 Exams You are here

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Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

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