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Last updated at May 12, 2021 by Teachoo

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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

Ex 2.2

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

Ex 2.2, 3 Deleted for CBSE Board 2022 Exams

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

Ex 2.2, 5 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 6 Deleted for CBSE Board 2022 Exams

Ex 2.2, 7 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 8 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 9 Deleted for CBSE Board 2022 Exams

Ex 2.2, 10 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 11 Deleted for CBSE Board 2022 Exams

Ex 2.2, 12 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 13 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 14 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 15 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 16 Deleted for CBSE Board 2022 Exams

Ex 2.2, 17 Deleted for CBSE Board 2022 Exams

Ex 2.2, 18 Important Deleted for CBSE Board 2022 Exams

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

Ex 2.2, 20 (MCQ) Important Deleted for CBSE Board 2022 Exams

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

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

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About the Author

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.