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Chapter 2 Class 12 Inverse Trigonometric Functions
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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

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