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Ex 2.2, 2 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 3 Deleted for CBSE Board 2022 Exams
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Ex 2.2, 9 Deleted for CBSE Board 2022 Exams
Ex 2.2, 10 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 11 Deleted for CBSE Board 2022 Exams
Ex 2.2, 12 Important Deleted for CBSE Board 2022 Exams
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Ex 2.2, 14 Important Deleted for CBSE Board 2022 Exams
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Ex 2.2, 18 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 20 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 21 (MCQ) Deleted for CBSE Board 2022 Exams
Last updated at May 12, 2021 by Teachoo
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 Solving cosβ1 ((π β π²π)/(π + π²π)) (Using 2 sin x cos x = sin 2x) cosβ1 ((π β π²π)/(π + π²π)) Putting y = tan πΆ = 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) = cosβ1 (ππππ πΆ β ππππ πΆ) = cosβ1 (πππβ‘ππΆ ) = 2Ξ± Hence, cosβ1 ((π β ππ)/(π + ππ)) = 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±