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Ex 2.2
Ex 2.2, 2 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 3 Deleted for CBSE Board 2023 Exams You are here
Ex 2.2, 4 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 5 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 6 Deleted for CBSE Board 2023 Exams
Ex 2.2, 7 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 8 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 9 Deleted for CBSE Board 2023 Exams
Ex 2.2, 10 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 11 Deleted for CBSE Board 2023 Exams
Ex 2.2, 12 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 13 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 14 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 15 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 16 Deleted for CBSE Board 2023 Exams
Ex 2.2, 17 Deleted for CBSE Board 2023 Exams
Ex 2.2, 18 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 19 (MCQ) Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 20 (MCQ) Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 21 (MCQ) Deleted for CBSE Board 2023 Exams
Last updated at March 16, 2023 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