Misc 11 - Chapter 2 Class 12 Inverse Trigonometry - tan-1 - Not clear how to approach

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

Transcript

Misc 11 Prove tan-1 ((√(1 + x) − √(1 − x))/(√(1 + x) + √(1 − x))) = π/4 - 1/2 cos-1 x, - 1/√2 ≤ x ≤ 1 [Hint: put x = cos 2θ] Taking L.H.S. tan-1 ((√(1 + x) − √(1 − x))/(√(1 + x) + √(1 − x))) Put x = cos 2𝜃 = tan-1 ((√(1 + cos 2θ) − √(1 − cos 2θ))/(√(1 + cos 2θ) + √(1 − cos 2θ))) = tan-1 ((√(2 cos^2 θ) − √(2 sin^2 θ))/(√(2 cos^2 θ) + √(2 sin^2 θ))) = tan-1 ((√2 cos θ − √2 sin θ)/(√2 cos θ + √(2 ) sin θ)) = tan-1 (( √2 (cos θ − sin θ))/(√2 (cos θ + sin θ))) = tan-1 (( (cos θ − sin θ))/( (cos θ + sin θ))) Dividing by cos θ = tan-1 ((cos⁡𝜃/cos⁡𝜃 − sin⁡𝜃/cos⁡𝜃 )/(cos⁡𝜃/cos⁡𝜃 + sin⁡𝜃/cos⁡𝜃 )) = tan-1 ((1 − tan θ)/(1 + tan θ)) = tan-1 ((1 − tan θ)/(1 + 1 . tan θ)) = tan-1 ((tan⁡〖 𝜋/4〗 − tan⁡θ)/(1 + tan⁡〖𝜋/4 .〖 tan〗⁡θ 〗 )) = tan-1 (tan (𝜋/4 −θ)) = 𝜋/4 − θ We know x = cos 2𝜃 cos-1 x = 2θ 1/2 cos-1 x = θ Hence, tan-1 ((√(1 + x) − √(1 − x))/(√(1 + x) + √(1 − x))) = 𝜋/4 – θ = 𝜋/4 – 1/2 cos-1x

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
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.