Miscellaneous
Misc. 2 Important
Misc. 3 Deleted for CBSE Board 2023 Exams
Misc. 4 Important Deleted for CBSE Board 2023 Exams
Misc. 5 Deleted for CBSE Board 2023 Exams
Misc. 6 Deleted for CBSE Board 2023 Exams
Misc. 7 Important Deleted for CBSE Board 2023 Exams
Misc. 8 Important Deleted for CBSE Board 2023 Exams
Misc. 9 Important Deleted for CBSE Board 2023 Exams
Misc. 10 Important Deleted for CBSE Board 2023 Exams
Misc. 11 Important Deleted for CBSE Board 2023 Exams
Misc 12 Important Deleted for CBSE Board 2023 Exams
Misc. 13 Important Deleted for CBSE Board 2023 Exams
Misc. 14 Deleted for CBSE Board 2023 Exams
Misc 15 (MCQ) Important Deleted for CBSE Board 2023 Exams You are here
Misc 16 (MCQ) Important Deleted for CBSE Board 2023 Exams
Misc 17 (MCQ) Deleted for CBSE Board 2023 Exams
Miscellaneous
Last updated at Aug. 9, 2021 by Teachoo
Misc 15 sin(tanβ1 x), |π₯| < 1 is equal to (A) π₯/β(1 β π₯2) (B) 1/β(1 β π₯2) (C) 1/β(1 + π₯2) (D) π₯/β(1 + π₯2) Let a = tanβ1 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‘ππ2 a) We convert tanβ1 to sinβ1 sec a = β(1+π₯2) 1/cosβ‘π = β(1+π₯2) 1/β(1 + π₯^2 ) = cosβ‘π πππβ‘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(γ1 + π₯γ^2 ) = π₯/β(γ1 + π₯γ^2 ) sin a = π₯/β(γ1 + π₯γ^2 ) a = sinβ1 (π₯/β(γ1 + π₯γ^2 )) Now solving sin(tanβ1 x) = sin (a) = sin ("sinβ1 " (π/β(γπ + πγ^π ))) = π₯/β(γ1 + π₯γ^2 ) Hence, D is the correct answer