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

Ques 10 (MCQ) - sin (tan^-1 x), where |x| < 1, is equal to - Teachoo - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

part 2 - Question 10 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 3 - Question 10 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 4 - Question 10 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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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

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 and Computer Science at Teachoo