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

Transcript

Misc 9 Prove tan−1 √x = 1/2 cos−1 ((1 − x)/(1 + x)), x ∈ [0, 1] Taking R.H.S. 1/2 cos−1 ((1 − x)/(1 + x)) Putting x = tan2 θ = 1/2 cos−1 ((1 − tan2θ)/(1 + tan2θ)) = 1/2 cos−1 ((1 − (sin2 θ)/(cos2 θ))/(1 + (sin2 θ)/(cos2 θ))) = 1/2 cos−1 (((cos2 θ − sin2 θ)/(cos2 θ))/((cos2 θ + sin2 θ)/(cos2 θ))) = 1/2 cos−1 ((cos2 θ − sin2 θ)/(cos2 θ + sin2 θ) ) = 1/2 cos−1 ((𝐜𝐨𝐬𝟐 𝛉 − 𝐬𝐢𝐧𝟐 𝛉)/1 ) = 1/2 cos−1 (cos 2𝛉) = 1/2 × 2θ = θ We assumed that x = tan2 θ √𝑥 = tan θ tan -1 √𝑥 = θ (Using cos2 θ + sin2 θ = 1) (Using cos 2θ = cos2 θ – sin2 θ) Hence, 1/2 cos−1 ((1 − x)/(1 + x)) = θ 1/2 cos−1 ((1 − x)/(1 + x)) = tan−1 √x Hence, R.H.S. = L.H.S. Hence Proved

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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.