Misc 8 - Prove tan-1 1/5 + tan-1 1/7 + tan-1 1/3 + tan-1 1/8

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Misc. 8 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2

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Misc. 8 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3

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

Transcript

Misc 8 Prove that tanโˆ’1 1/5 + tanโˆ’1 1/7 + tanโˆ’1 1/3 + tanโˆ’1 1/8 = ๐œ‹/4 We know that tanโˆ’1 x + tanโˆ’1 y = tanโˆ’1 ((๐ฑ + ๐ฒ )/(๐Ÿ โˆ’ ๐ฑ๐ฒ)) tanโˆ’1 ๐Ÿ/๐Ÿ“ + tanโˆ’1 ๐Ÿ/๐Ÿ• = tanโˆ’1 (1/5 + 1/7)/(1โˆ’ 1/5 ร— 1/7) = tanโˆ’1 ((7 + 5)/(5(7)))/( (35 โˆ’ 1)/35 ) = tanโˆ’1 (6/17) tanโˆ’1 ๐Ÿ/๐Ÿ‘ + tanโˆ’1 ๐Ÿ/๐Ÿ– = tanโˆ’1 (1/3 + 1/8)/(1โˆ’ 1/3 ร— 1/8) = tanโˆ’1 ( (8 + 3)/(3(8)))/( (24 โˆ’ 1)/24) = tan โˆ’ 1 (11/23) Solving L.H.S tanโˆ’1 1/5 + tanโˆ’1 1/7 + tanโˆ’1 1/3 + tanโˆ’1 1/8 = ("tanโˆ’1 " 1/5 " + tanโˆ’1 " 1/7) + ("tanโˆ’1 " 1/3 " + tanโˆ’1 " 1/8) = tan-1 (๐Ÿ”/๐Ÿ๐Ÿ•) + tanโˆ’1 (๐Ÿ๐Ÿ/๐Ÿ๐Ÿ‘) = tanโˆ’1 (6/17 + 11/23)/(1 โˆ’ 6/17 ร— 11/23) We know that tanโˆ’1 x + tanโˆ’1 y = tanโˆ’1 ((๐’™ + ๐’š )/(๐Ÿ โˆ’ ๐’™๐’š)) Replacing x by 6/17 and y by 11/23 = tanโˆ’1 ( (6(23) + 11(17))/(17(23)))/(1 โˆ’ 66/391) = tanโˆ’1 ( (138 + 187)/391)/((391 โˆ’ 66)/391) = tanโˆ’1 ( 325/391)/(325/391) = tanโˆ’1 1 = tanโˆ’1 ("tan " ๐…/๐Ÿ’) = ฯ€/4 = R.H.S Hence proved (As tan ๐œ‹/4 = 1)

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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.