
Ex 2.2
Ex 2.2, 2 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 3 Deleted for CBSE Board 2022 Exams You are here
Ex 2.2, 4 Important Deleted for CBSE Board 2022 Exams
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 Aug. 11, 2021 by Teachoo
Ex 2.2, 3 Prove tan−1 2/11 + tan−1 7/24 = tan−1 1/2 Solving L.H.S. tan-1 2/11 + tan−1 7/24 = "tan−1" ((2/11 + 7/24)/(1− 2/11 × 7/24)) We know that tan-1 x + tan−1 y = tan−1 ((𝒙+𝒚 )/(𝟏 −𝒙𝒚)) Replace x by 2/11 and y by 7/24= tan-1 (((24 × 2 + 7 × 11)/(24 × 11))/((11 × 12 − 7)/(11 × 12))) = tan-1 (((48 + 77)/(24 × 11))/((132 − 7)/(11 × 12))) = tan-1 ((125/(24 × 11))/(125/(11 × 12))) = tan-1 (125/(24 × 11) × (11 × 12)/125) = tan−1 (1/2) = R.H.S. Hence. R.H.S. = L.H.S. Hence Proved