Chapter 2 Class 12 Inverse Trigonometric Functions
Concept wise

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Misc 13 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) 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.