Ex 2.2

Ex 2.2,1
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Ex 2.2, 2 Important Deleted for CBSE Board 2022 Exams

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Ex 2.2, 4 Important Deleted for CBSE Board 2022 Exams You are here

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Ex 2.2, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams

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Ex 2.2, 21 (MCQ) Deleted for CBSE Board 2022 Exams

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

Serial order wise

Last updated at May 12, 2021 by Teachoo

Ex 2.2, 4 Prove 2tanโ1 1/2 + tanโ1 1/7 = tanโ1 31/17 Value of 2tanโ1 ๐/๐ We know that 2tanโ1x = tanโ1 ((๐๐ฑ )/( ๐ โ ๐ฑ^๐ )) Replacing x with 1/2 2tanโ1 1/2 = tanโ1 (2 ร 1/2)/(1 โ (1/2)2) = tanโ1 (1/(1 โ 1/4)) = tanโ1 (1/((4 โ 1)/4)) = tanโ1 (1/(3/4)) = tanโ1 (๐/๐) Solving L.H.S. 2tanโ1 1/2 + tanโ1 1/7 Putting value of 2tanโ1 1/2 = tanโ1 4/3 + tanโ1 1/7 = tanโ1 (1/(1 โ 1/4)) = tanโ1 (1/((4 โ 1)/4)) = tanโ1 (1/(3/4)) = tanโ1 (๐/๐) Solving L.H.S. 2tanโ1 1/2 + tanโ1 1/7 Putting value of 2tanโ1 1/2 = tanโ1 4/3 + tanโ1 1/7 Using tanโ1x + tanโ1y = tanโ1 ((๐ + ๐ )/( ๐โ ๐๐)) Replacing x by 4/3 and y by 1/(7 )= tanโ1 ((๐/๐ + ๐/๐ )/( ๐โ ๐/๐ ร ๐/๐)) = tanโ1 (((4 ร 7 +3 ร 1 )/( 7 ร 3) )/( (7 ร 3 โ 4)/(7 ร 3))) = tanโ1 (((28 + 3 )/( 21) )/( ( 21 โ 4)/21)) = tanโ1 ((31/( 21) )/(17/21)) = tanโ1 (31/21ร21/17) = tanโ1 (๐๐/๐๐) = R.H.S. Hence, L.H.S. = R.H.S. Hence Proved