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Ex 2.2, 4 - Prove 2tan-1 1/2 + tan-1 1/7 = tan-1 31/17 - Ex 2.2

Ex 2.2, 4 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Ex 2.2, 4 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3

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Transcript

Ex 2.2, 4 Prove 2tanโˆ’1 1/2 + tanโˆ’1 1/7 = tanโˆ’1 31/17 Value of 2tanโˆ’1 ๐Ÿ/๐Ÿ We know that 2tanโˆ’1x = tanโˆ’1 ((๐Ÿ๐ฑ )/( ๐Ÿ โˆ’ ๐ฑ^๐Ÿ )) Replacing x with 1/2 2tanโˆ’1 1/2 = tanโˆ’1 (2 ร— 1/2)/(1 โˆ’ (1/2)2) = tanโˆ’1 (1/(1 โˆ’ 1/4)) = tanโˆ’1 (1/((4 โˆ’ 1)/4)) = tanโˆ’1 (1/(3/4)) = tanโˆ’1 (๐Ÿ’/๐Ÿ‘) Solving L.H.S. 2tanโˆ’1 1/2 + tanโˆ’1 1/7 Putting value of 2tanโˆ’1 1/2 = tanโˆ’1 4/3 + tanโˆ’1 1/7 = tanโˆ’1 (1/(1 โˆ’ 1/4)) = tanโˆ’1 (1/((4 โˆ’ 1)/4)) = tanโˆ’1 (1/(3/4)) = tanโˆ’1 (๐Ÿ’/๐Ÿ‘) Solving L.H.S. 2tanโˆ’1 1/2 + tanโˆ’1 1/7 Putting value of 2tanโˆ’1 1/2 = tanโˆ’1 4/3 + tanโˆ’1 1/7 Using tanโˆ’1x + tanโˆ’1y = tanโˆ’1 ((๐’™ + ๐’š )/( ๐Ÿโˆ’ ๐’™๐’š)) Replacing x by 4/3 and y by 1/(7 )= tanโˆ’1 ((๐Ÿ’/๐Ÿ‘ + ๐Ÿ/๐Ÿ• )/( ๐Ÿโˆ’ ๐Ÿ’/๐Ÿ‘ ร— ๐Ÿ/๐Ÿ•)) = tanโˆ’1 (((4 ร— 7 +3 ร— 1 )/( 7 ร— 3) )/( (7 ร— 3 โˆ’ 4)/(7 ร— 3))) = tanโˆ’1 (((28 + 3 )/( 21) )/( ( 21 โˆ’ 4)/21)) = tanโˆ’1 ((31/( 21) )/(17/21)) = tanโˆ’1 (31/21ร—21/17) = tanโˆ’1 (๐Ÿ‘๐Ÿ/๐Ÿ๐Ÿ•) = R.H.S. Hence, L.H.S. = R.H.S. Hence Proved

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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.