

Ex 2.2
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
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