Misc 15 - sin (tan-1 x) is equal to  - Chapter 2 Inverse NCERT - Changing of trignometric variables and then applying formula

  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
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Misc 15 sin(tan-1 x), |๐‘ฅ| < 1 is equal to (A) ๐‘ฅ/โˆš(1 โˆ’ ๐‘ฅ2) (B) 1/โˆš(1 โˆ’ ๐‘ฅ2) (C) 1/โˆš(1 + ๐‘ฅ2) (D) ๐‘ฅ/โˆš(1 + ๐‘ฅ2) 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 = โˆš(1+๐‘ก๐‘Ž๐‘›2 a) sec a = โˆš(1+๐‘ฅ2) 1/cosโก๐‘Ž = โˆš(1+๐‘ฅ2) 1/โˆš(1 + ๐‘ฅ^2 ) = cosโก๐‘Ž cosโก๐‘Ž = 1/โˆš(1 + ๐‘ฅ^2 ) We know that sin2 a = 1 โ€“ cos2 a sin a = โˆš("1 โ€“ cos2 a" ) sin a = โˆš("1 โ€“" (1/โˆš(1 + ๐‘ฅ^2 ) " " )^2 ) sin a = โˆš("1 โ€“" 1/(1 + ๐‘ฅ2)) sin a = โˆš((1 + ๐‘ฅ2 โˆ’ 1)/(1 + ๐‘ฅ2)) = โˆš((๐‘ฅ2 )/(1 + ๐‘ฅ2)) = โˆš(๐‘ฅ^2 )/โˆš(ใ€–1 + ๐‘ฅใ€—^2 ) = ๐‘ฅ/โˆš(ใ€–1 + ๐‘ฅใ€—^2 ) sin a = ๐‘ฅ/โˆš(ใ€–1 + ๐‘ฅใ€—^2 ) a = sin-1 (๐‘ฅ/โˆš(ใ€–1 + ๐‘ฅใ€—^2 )) Now solving sin(tan-1 x) = sin (a) = sin (sin-1 (๐‘ฅ/โˆš(ใ€–1 + ๐‘ฅใ€—^2 ))) = ๐‘ฅ/โˆš(ใ€–1 + ๐‘ฅใ€—^2 ) Hence D is the correct answer

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.