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

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

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

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

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
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.