Last updated at Dec. 8, 2016 by Teachoo

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Ex 2.2, 13 tan 1/2 [sin-1 2π₯/(1 + π₯2) + cos-1 (1 β π¦2)/(1 + π¦2) ] We solve sin-1 2π₯/(1 + π₯2) & cos-1 ((1 β π¦2)/(1 + π¦2)) separately Solving sin-1 ππ/(π + ππ) sin-1 2π₯/(1 + π₯2) Putting x = tan ΞΈ = sin-1 ((2 tanβ‘π)/(1 + π‘ππ2 π)) = sin-1 ((2 tanβ‘π)/(π ππ2 π)) = sin-1 ((2 sinβ‘ΞΈ/cosβ‘ΞΈ )/(1/cos2β‘ΞΈ )) = sin-1 (2 sinβ‘ΞΈ/cosβ‘ΞΈ Γ cos2β‘ΞΈ/1) = sin-1 (2 sinβ‘ΞΈ cosβ‘ΞΈ ) = sin-1 (sin 2ΞΈ) = 2ΞΈ β΄ sin-1 2π₯/(1 + π₯2) = 2ΞΈ Now solving cos-1 ((π β π²π)/(π + π²π)) cos-1 ((1 β y2)/(1 + y2)) Putting y = tan πΌ = cosβ1 ((1 βγ tan2γβ‘πΌ)/(1 +γ tan2γβ‘πΌ )) = cosβ1 ((1 βγ tan2γβ‘πΌ)/sec2β‘πΌ ) = cosβ1 ((1 β (π ππ2 πΌ )/(πππ 2 πΌ))/((1 )/(πππ 2 πΌ))) = cos-1 (( (πππ 2 πΌ β π ππ2 πΌ )/(πππ 2 πΌ))/((1 )/(πππ 2 πΌ))) = cos-1 ((πππ 2 πΌ β π ππ2 πΌ )/(πππ 2 πΌ) Γ (πππ 2 πΌ)/1) = cos-1 (πππ 2 πΌ β π ππ2 πΌ) = cos-1 (cosβ‘2πΌ ) Hence, cos-1 ((1 β π¦2)/(1 + π¦2)) = 2Ξ± Now, we have sin-1 2π₯/(1 + π₯2) = 2ΞΈ & cos-1 ((1 β π¦2)/(1 + π¦2)) = 2Ξ± Solving tan 1/2 [sin-1 2π₯/(1 + π₯2) + cos-1 (1 β π¦2)/(1 + π¦2) ] Putting values = tan 1/2 [2ΞΈ + 2Ξ±] = tan 1/2 [2(ΞΈ + Ξ±)] = tan (ΞΈ + Ξ±) = (tanβ‘ΞΈ + tanβ‘πΌ)/(γ1 β tanγβ‘ΞΈ tanβ‘πΌ ) = (π₯ + π¦)/(1 β π₯π¦) Hence, tan 1/2 [sin-1 2π₯/(1 + π₯2) + cos-1 (1 β π¦2)/(1 + π¦2) ] = (π₯ + π¦)/(1 β π₯π¦)

Chapter 2 Class 12 Inverse Trigonometric Functions

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.