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

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

## (a) x/β(1 - x 2 ) Β Β  (b) 1/β(1 - x 2 ) (c) 1/β(1 + x 2 )Β  Β (d) x/β(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 (tanβ1x), where |x| < 1, is equal to: (a) π₯/β(1 β π₯^2 ) (b) 1/β(1 βγ π₯γ^2 ) (c) 1/β(1 + π₯^2 ) (d) π₯/β(1 + π₯^2 ) Let a = tanβ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 = β(1+π‘ππ2 a) sec a = β(1+π₯2) 1/cosβ‘π = β(1+π₯2) 1/β(1 + π₯^2 ) = cosβ‘π πππβ‘π = π/β(π + π^π ) We know that sin a = β("1 β cos2 a" ) sin a = β("1 β" (1/β(1 + π₯^2 ))^2 ) sin a = β("1 β" 1/(1 + π₯2)) sin a = β((1 + π₯2 β 1)/(1 + π₯2)) = β((π₯2 )/(1 + π₯2)) = β(π₯^2 )/β(γ1 + π₯γ^2 ) = π₯/β(γ1 + π₯γ^2 ) sin a = π₯/β(γ1 + π₯γ^2 ) a = sinβ1 (π/β(γπ + πγ^π )) Now solving sin(tanβ1 x) = sin (a) = sin ("sinβ1 " (π/β(γπ + πγ^π ))) = π₯/β(γ1 + π₯γ^2 ) So, the correct answer is (d)