1. Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
  2. Serial order wise
  3. Examples

Example 4 - Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

Last updated at May 12, 2021 by

Example 4 - Show tan-1 1/2 + tan-1 2/11 = tan-1 3/4 - Examples

Example 4 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2

  1. Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
  2. Serial order wise

    3. Examples

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Transcript

Example 4 Show that tan-1 1/2 + tan-1 2/11 = tan-1 3/4 Taking L.H.S tan-1 1/2 + tan-1 2/11 = tan-1 (1/2 + 2/11)/(1− 1/2 × 2/11) = tan-1 (1/2 + 2/11)/(1− 1/11) = tan-1 ((1(11) + 2(2))/(2 × 11) )/((1(11) − 1)/11) = tan-1 ((11 + 4)/(2 × 11) )/((11 − 1)/11) = tan-1 (15/(2 × 11) )/(10/11) = tan-1 (15/(2 × 11) × 11/10) = tan-1 3/4 We know that tan-1 x + tan-1 y = tan-1 ((𝒙 + 𝒚 )/(𝟏 −𝒙𝒚)) Replacing x by 1/2 and y by 2/11 = R.H.S. Thus L.H.S = R.H.S Hence proved

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
Serial order wise

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.