Solve sin (tan^-1 x) - Inverse Trigonometry - Teachoo - Miscellaneous

Misc. 15 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Misc. 15 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3 Misc. 15 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 4

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

Transcript

Misc 15 sin(tanβˆ’1 x), |π‘₯| < 1 is equal to (A) π‘₯/√(1 βˆ’ π‘₯2) (B) 1/√(1 βˆ’ π‘₯2) (C) 1/√(1 + π‘₯2) (D) π‘₯/√(1 + π‘₯2) Let a = tanβˆ’1 x tan a = x We need to find sin a. For this first we calculate sec a and cos a We know that sec2 a = 1 + tan2 a sec a = √(1+π‘‘π‘Žπ‘›2 a) We convert tanβˆ’1 to sinβˆ’1 sec a = √(1+π‘₯2) 1/cosβ‘π‘Ž = √(1+π‘₯2) 1/√(1 + π‘₯^2 ) = cosβ‘π‘Ž 𝒄𝒐𝒔⁑𝒂 = 𝟏/√(𝟏 + 𝒙^𝟐 ) We know that sin a = √("1 – cos2 a" ) sin a = √("1 –" (1/√(1 + π‘₯^2 ))^2 ) sin a = √("1 –" 1/(1 + π‘₯2)) sin a = √((1 + π‘₯2 βˆ’ 1)/(1 + π‘₯2)) = √((π‘₯2 )/(1 + π‘₯2)) = √(π‘₯^2 )/√(γ€–1 + π‘₯γ€—^2 ) = π‘₯/√(γ€–1 + π‘₯γ€—^2 ) sin a = π‘₯/√(γ€–1 + π‘₯γ€—^2 ) a = sinβˆ’1 (π‘₯/√(γ€–1 + π‘₯γ€—^2 )) Now solving sin(tanβˆ’1 x) = sin (a) = sin ("sinβˆ’1 " (𝒙/√(γ€–πŸ + 𝒙〗^𝟐 ))) = π‘₯/√(γ€–1 + π‘₯γ€—^2 ) Hence, D is the correct answer

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.