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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

Last updated at Dec. 16, 2024 by Teachoo

sin (tan−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

Transcript

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)
  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

    3. CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

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CBSE Class 12 Sample Paper for 2021 Boards →
Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

About the Author

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