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Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

sin (tanβˆ’1x), where |x| < 1, is equal to:

(a) x/√(1 - x 2 )    (b) 1/√(1 - x 2 )
(c) 1/√(1 + x 2 )   (d) x/√(1 + x 2 )

This question is inspired from Misc 15 (MCQ) - Chapter 2 Class 12 - Inverse Trigonometric Functions

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Question 10 sin (tanβˆ’1x), where |x| < 1, is equal to: (a) π‘₯/√(1 βˆ’ π‘₯^2 ) (b) 1/√(1 βˆ’γ€– π‘₯γ€—^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 (𝒙/√(γ€–πŸ + 𝒙〗^𝟐 )) Now solving sin(tanβˆ’1 x) = sin (a) = sin ("sinβˆ’1 " (𝒙/√(γ€–πŸ + 𝒙〗^𝟐 ))) = π‘₯/√(γ€–1 + π‘₯γ€—^2 ) So, the correct answer is (d)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.