Last updated at May 29, 2018 by Teachoo

Transcript

Misc 15 sin(tan-1 x), |๐ฅ| < 1 is equal to (A) ๐ฅ/โ(1 โ ๐ฅ2) (B) 1/โ(1 โ ๐ฅ2) (C) 1/โ(1 + ๐ฅ2) (D) ๐ฅ/โ(1 + ๐ฅ2) Let a = tan-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 = โ(1+๐ก๐๐2 a) sec a = โ(1+๐ฅ2) 1/cosโก๐ = โ(1+๐ฅ2) 1/โ(1 + ๐ฅ^2 ) = cosโก๐ cosโก๐ = 1/โ(1 + ๐ฅ^2 ) We know that sin2 a = 1 โ cos2 a sin a = โ("1 โ cos2 a" ) sin a = โ("1 โ" (1/โ(1 + ๐ฅ^2 ) " " )^2 ) sin a = โ("1 โ" 1/(1 + ๐ฅ2)) sin a = โ((1 + ๐ฅ2 โ 1)/(1 + ๐ฅ2)) = โ((๐ฅ2 )/(1 + ๐ฅ2)) = โ(๐ฅ^2 )/โ(ใ1 + ๐ฅใ^2 ) = ๐ฅ/โ(ใ1 + ๐ฅใ^2 ) sin a = ๐ฅ/โ(ใ1 + ๐ฅใ^2 ) a = sin-1 (๐ฅ/โ(ใ1 + ๐ฅใ^2 )) Now solving sin(tan-1 x) = sin (a) = sin (sin-1 (๐ฅ/โ(ใ1 + ๐ฅใ^2 ))) = ๐ฅ/โ(ใ1 + ๐ฅใ^2 ) Hence D is the correct answer

Chapter 2 Class 12 Inverse Trigonometric Functions

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.