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Last updated at Feb. 13, 2020 by Teachoo

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We know that sin"β"1 x + cos"β"1 x = π/2 sin"β"1 x = π/2 β cos"β"1 x Replacing x by (π₯ β 1)/(π₯ β 2) and y by ((π₯ + 1)/(π₯ + 2)) = tanβ1 [((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))] = tanβ1 [((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 = tanβ1 [((x β 1) (x + 2) + (x + 1)(x β 2))/(π₯2 β 22 β[π₯2 β 12]) ] = tanβ1 [(π₯ (x + 2) β 1 (x + 2) + x (x β 2) + 1 (x β 2))/(π₯2 β 4 β π₯2 + 1) ] = tanβ1 [(π₯2 + 2π₯ β π₯ β 2 + π₯2 β 2π₯ + π₯ β 2 )/(π₯2 β π₯2 β 4 + 1) ] = tanβ1 [(2x2 β4)/(β3)] Therefore, tanβ1 ((π± β π)/(π± β π)) + tanβ1 ((π± + π)/(π± + π)) = tanβ1 [(ππ±π βπ)/(βπ)] Given tanβ1 (x β 1)/(x β 2) + tanβ1 (x + 1)/(x + 2) = π/4 Putting values tanβ1 [(2x2 β4)/(β3)] = Ο/4 (2x2 β4)/(β3) = tan Ο/4 (2x2 β4)/(β3) = 1 2x2 β 4 = β3 2x2 = β3 + 4 2x2 = 1 x2 = 1/2 x = Β± 1/β2 Thus, x = Β± π/βπ

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Chapter 2 Class 12 Inverse Trigonometric Functions

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.