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Formulae based

Inverse Trigonometry Formulas

Example 3 (i) Important Deleted for CBSE Board 2023 Exams

Ex 2.2,1 Deleted for CBSE Board 2023 Exams

Ex 2.2, 2 Important Deleted for CBSE Board 2023 Exams

Example 8 Deleted for CBSE Board 2023 Exams

Ex 2.2, 12 Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 14 Important Deleted for CBSE Board 2023 Exams

Example 4 Deleted for CBSE Board 2023 Exams

Ex 2.2, 3 Deleted for CBSE Board 2023 Exams

Misc. 8 Important Deleted for CBSE Board 2023 Exams

Ex 2.2, 4 Important Deleted for CBSE Board 2023 Exams

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

Example 13 Important Deleted for CBSE Board 2023 Exams

Misc. 14 Deleted for CBSE Board 2023 Exams

Misc 17 (MCQ) Deleted for CBSE Board 2023 Exams

Misc. 13 Important Deleted for CBSE Board 2023 Exams

Example 10 Important Deleted for CBSE Board 2023 Exams

Misc. 3 Deleted for CBSE Board 2023 Exams

Misc. 4 Important Deleted for CBSE Board 2023 Exams

Misc 12 Important Deleted for CBSE Board 2023 Exams

Misc 16 (MCQ) Important Deleted for CBSE Board 2023 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