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Misc 9 - Prove tan-1 root x  = 1/2 cos-1 (1 - x)/(1 + x) - Not clear how to approach

  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
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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) × (cos2 θ)/(cos2 θ + sin2 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 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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