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

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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.