Changing of trignometric variables and then applying formula

Chapter 2 Class 12 Inverse Trigonometric Functions
Concept wise

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Misc 15 sin(tanβ1 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 = 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) We convert tanβ1 to sinβ1 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1 + π₯γ^2 )) 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 ) Hence, D is the correct answer