Ex 2.2, 13 - Inverse Trigonometry - tan 1/2 [sin-1 2x/1+x2 - Ex 2.2

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

Transcript

Ex 2.2, 13 tan 1/2 [sin-1 2π‘₯/(1 + π‘₯2) + cos-1 (1 βˆ’ 𝑦2)/(1 + 𝑦2) ] We solve sin-1 2π‘₯/(1 + π‘₯2) & cos-1 ((1 βˆ’ 𝑦2)/(1 + 𝑦2)) separately Solving sin-1 πŸπ’™/(𝟏 + π’™πŸ) sin-1 2π‘₯/(1 + π‘₯2) Putting x = tan ΞΈ = sin-1 ((2 tanβ‘πœƒ)/(1 + π‘‘π‘Žπ‘›2 πœƒ)) = sin-1 ((2 tanβ‘πœƒ)/(𝑠𝑒𝑐2 πœƒ)) = sin-1 ((2 sin⁑θ/cos⁑θ )/(1/cos2⁑θ )) = sin-1 (2 sin⁑θ/cos⁑θ Γ— cos2⁑θ/1) = sin-1 (2 sin⁑θ cos⁑θ ) = sin-1 (sin 2ΞΈ) = 2ΞΈ ∴ sin-1 2π‘₯/(1 + π‘₯2) = 2ΞΈ Now solving cos-1 ((𝟏 βˆ’ 𝐲𝟐)/(𝟏 + 𝐲𝟐)) cos-1 ((1 βˆ’ y2)/(1 + y2)) Putting y = tan 𝛼 = cosβˆ’1 ((1 βˆ’γ€– tan2〗⁑𝛼)/(1 +γ€– tan2〗⁑𝛼 )) = cosβˆ’1 ((1 βˆ’γ€– tan2〗⁑𝛼)/sec2⁑𝛼 ) = cosβˆ’1 ((1 βˆ’ (𝑠𝑖𝑛2 𝛼 )/(π‘π‘œπ‘ 2 𝛼))/((1 )/(π‘π‘œπ‘ 2 𝛼))) = cos-1 (( (π‘π‘œπ‘ 2 𝛼 βˆ’ 𝑠𝑖𝑛2 𝛼 )/(π‘π‘œπ‘ 2 𝛼))/((1 )/(π‘π‘œπ‘ 2 𝛼))) = cos-1 ((π‘π‘œπ‘ 2 𝛼 βˆ’ 𝑠𝑖𝑛2 𝛼 )/(π‘π‘œπ‘ 2 𝛼) Γ— (π‘π‘œπ‘ 2 𝛼)/1) = cos-1 (π‘π‘œπ‘ 2 𝛼 βˆ’ 𝑠𝑖𝑛2 𝛼) = cos-1 (cos⁑2𝛼 ) Hence, cos-1 ((1 βˆ’ 𝑦2)/(1 + 𝑦2)) = 2Ξ± Now, we have sin-1 2π‘₯/(1 + π‘₯2) = 2ΞΈ & cos-1 ((1 βˆ’ 𝑦2)/(1 + 𝑦2)) = 2Ξ± Solving tan 1/2 [sin-1 2π‘₯/(1 + π‘₯2) + cos-1 (1 βˆ’ 𝑦2)/(1 + 𝑦2) ] Putting values = tan 1/2 [2ΞΈ + 2Ξ±] = tan 1/2 [2(ΞΈ + Ξ±)] = tan (ΞΈ + Ξ±) = (tan⁑θ + tan⁑𝛼)/(γ€–1 βˆ’ tan〗⁑θ tan⁑𝛼 ) = (π‘₯ + 𝑦)/(1 βˆ’ π‘₯𝑦) Hence, tan 1/2 [sin-1 2π‘₯/(1 + π‘₯2) + cos-1 (1 βˆ’ 𝑦2)/(1 + 𝑦2) ] = (π‘₯ + 𝑦)/(1 βˆ’ π‘₯𝑦)

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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.