Misc 3 - Prove 2 sin-1 3/5 = tan-1 24/7 - Chapter 2 Inverse

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Misc. 3 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2

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Misc. 3 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3 Misc. 3 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 4

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

Transcript

Misc 3 Prove 2 sin-1 3/5 = tan-1 24/7 We need to convert LHS in form tan-1 Converting sin-1 (πŸ‘/πŸ“) to tan-1 Let x = sin-1 (3/5) sin x = 3/5 Now, cos x = √(1βˆ’π‘ π‘–π‘›2 π‘₯) = √(1 βˆ’ (3/5)^2 ) = √(1 βˆ’ 9/25) = √((25 βˆ’ 9)/25) = √(16/25) = 4/5 Thus, tan x = sin⁑π‘₯/cos⁑π‘₯ tan x = (3/5)/(4/5) tan x = 3/4 x = tan–1 πŸ‘/πŸ’ Solving L.H.S 2 sin–1 πŸ‘/πŸ“ = 2x = 2 tan-1 (3/4) Using 2tan-1 x = tan-1 (πŸπ’™/(𝟏 βˆ’ π’™πŸ)) = tan-1 (2(3/4)/(1 βˆ’ (3/4)2)) = tan-1 ((3/2)/(1 βˆ’ 9/16)) = tan-1 ((3/2)/( (16 βˆ’ 9)/16)) = tan-1 ((3/2)/( 7/16)) = tan-1 (3/2Γ—16/7) = tan-1 (πŸπŸ’/πŸ•) = R.H.S. Hence L.H.S. = R.H.S Hence proved

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Davneet Singh
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.