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Last updated at Feb. 13, 2020 by Teachoo

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Ex 2.2, 13 tan 1/2 [sinβ1 2π₯/(1 + π₯2) + cosβ1 (1 β π¦2)/(1 + π¦2) ] We solve sinβ1 2π₯/(1 + π₯2) & cosβ1 ((1 β π¦2)/(1 + π¦2)) separately Solving sinβ1 ππ/(π + ππ) sinβ1 2π₯/(1 + π₯2) Putting x = tan ΞΈ = sinβ1 ((2 tanβ‘π)/(π + ππππ π½)) = sinβ1 ((2 tanβ‘π)/(ππππ π½)) We write sin-1 & cosβ1 in terms of tan-1 When there is 1 + x2 , we put x = tan ΞΈ (Using 1 + tan2 x = sec2 x) = sinβ1 ((2 sinβ‘ΞΈ/cosβ‘ΞΈ )/(1/cos2β‘ΞΈ )) = sinβ1 (2 sinβ‘ΞΈ/cosβ‘ΞΈ Γcos2β‘ΞΈ/1) = sinβ1 (π πππβ‘π½ πππβ‘π½ ) = sinβ1 (sin 2ΞΈ) = 2ΞΈ β΄ sinβ1 2π₯/(1 + π₯2) = 2ΞΈ Now solving cosβ1 ((π β π²π)/(π + π²π)) (Using 2 sin x cos x = sin 2x) cosβ1 ((1 β y2)/(1 + y2)) Putting y = tan πΌ = cosβ1 ((1 βγ tan2γβ‘πΌ)/(π +γ πππ§πγβ‘πΆ )) = cosβ1 ((1 βγ tan2γβ‘πΌ)/π¬πππβ‘πΆ ) = cosβ1 ((1 β (π ππ2 πΌ )/(πππ 2 πΌ))/((1 )/(πππ 2 πΌ))) = cosβ1 (( (πππ 2 πΌ β π ππ2 πΌ )/(πππ 2 πΌ))/((1 )/(πππ 2 πΌ))) = cosβ1 ((πππ 2 πΌ β π ππ2 πΌ )/(πππ 2 πΌ)Γ(πππ 2 πΌ)/1) (Using 1 + tan2 x = sec2 x) = cosβ1 (ππππ πΆ β ππππ πΆ) = cosβ1 (πππβ‘ππΆ ) Hence, cosβ1 ((1 β π¦2)/(1 + π¦2)) = 2Ξ± Now, we have sinβ1 2π₯/(1 + π₯2) = 2ΞΈ & cosβ1 ((1 β π¦2)/(1 + π¦2)) = 2Ξ± Solving tan π/π ["sinβ1 " ππ/(π + ππ) " + cosβ1 " (π β ππ)/(π + ππ)] Putting values (Using cos 2x = cos2 x β sin2 x) = tan 1/2 [2ΞΈ + 2Ξ±] = tan (ΞΈ + Ξ±) = (tanβ‘ΞΈ + tanβ‘πΌ)/(γ1 β tanγβ‘ΞΈ tanβ‘πΌ ) = (π + π)/(π β ππ) Putting x = tan ΞΈ & y = tan Ξ±

Ex 2.2

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Chapter 2 Class 12 Inverse Trigonometric Functions

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.