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  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise

Transcript

Ex 2.2, 13 tan 1/2 [sinβˆ’1 2π‘₯/(1 + π‘₯2) + cosβˆ’1 (1 βˆ’ 𝑦2)/(1 + 𝑦2) ] We solve sinβˆ’1 2π‘₯/(1 + π‘₯2) & cosβˆ’1 ((1 βˆ’ 𝑦2)/(1 + 𝑦2)) separately Solving sinβˆ’1 πŸπ’™/(𝟏 + π’™πŸ) sinβˆ’1 2π‘₯/(1 + π‘₯2) Putting x = tan ΞΈ = sinβˆ’1 ((2 tanβ‘πœƒ)/(𝟏 + π’•π’‚π’πŸ 𝜽)) = sinβˆ’1 ((2 tanβ‘πœƒ)/(π’”π’†π’„πŸ 𝜽)) We write sin-1 & cosβˆ’1 in terms of tan-1 When there is 1 + x2 , we put x = tan ΞΈ (Using 1 + tan2 x = sec2 x) = sinβˆ’1 ((2 sin⁑θ/cos⁑θ )/(1/cos2⁑θ )) = sinβˆ’1 (2 sin⁑θ/cos⁑θ Γ—cos2⁑θ/1) = sinβˆ’1 (𝟐 π’”π’Šπ’β‘πœ½ π’„π’π’”β‘πœ½ ) = sinβˆ’1 (sin 2ΞΈ) = 2ΞΈ ∴ sinβˆ’1 2π‘₯/(1 + π‘₯2) = 2ΞΈ Now solving cosβˆ’1 ((𝟏 βˆ’ 𝐲𝟐)/(𝟏 + 𝐲𝟐)) (Using 2 sin x cos x = sin 2x) cosβˆ’1 ((1 βˆ’ y2)/(1 + y2)) Putting y = tan 𝛼 = cosβˆ’1 ((1 βˆ’γ€– tan2〗⁑𝛼)/(𝟏 +γ€– π­πšπ§πŸγ€—β‘πœΆ )) = cosβˆ’1 ((1 βˆ’γ€– tan2〗⁑𝛼)/𝐬𝐞𝐜𝟐⁑𝜢 ) = cosβˆ’1 ((1 βˆ’ (𝑠𝑖𝑛2 𝛼 )/(π‘π‘œπ‘ 2 𝛼))/((1 )/(π‘π‘œπ‘ 2 𝛼))) = cosβˆ’1 (( (π‘π‘œπ‘ 2 𝛼 βˆ’ 𝑠𝑖𝑛2 𝛼 )/(π‘π‘œπ‘ 2 𝛼))/((1 )/(π‘π‘œπ‘ 2 𝛼))) = cosβˆ’1 ((π‘π‘œπ‘ 2 𝛼 βˆ’ 𝑠𝑖𝑛2 𝛼 )/(π‘π‘œπ‘ 2 𝛼)Γ—(π‘π‘œπ‘ 2 𝛼)/1) (Using 1 + tan2 x = sec2 x) = cosβˆ’1 (π’„π’π’”πŸ 𝜢 βˆ’ π’”π’Šπ’πŸ 𝜢) = cosβˆ’1 (π’„π’π’”β‘πŸπœΆ ) Hence, cosβˆ’1 ((1 βˆ’ 𝑦2)/(1 + 𝑦2)) = 2Ξ± Now, we have sinβˆ’1 2π‘₯/(1 + π‘₯2) = 2ΞΈ & cosβˆ’1 ((1 βˆ’ 𝑦2)/(1 + 𝑦2)) = 2Ξ± Solving tan 𝟏/𝟐 ["sinβˆ’1 " πŸπ’™/(𝟏 + π’™πŸ) " + cosβˆ’1 " (𝟏 βˆ’ π’šπŸ)/(𝟏 + π’šπŸ)] Putting values (Using cos 2x = cos2 x – sin2 x) = tan 1/2 [2ΞΈ + 2Ξ±] = tan (ΞΈ + Ξ±) = (tan⁑θ + tan⁑𝛼)/(γ€–1 βˆ’ tan〗⁑θ tan⁑𝛼 ) = (𝒙 + π’š)/(𝟏 βˆ’ π’™π’š) Putting x = tan ΞΈ & y = tan Ξ±

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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.