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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
Ex 2.2, 4 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 5 Important Deleted for CBSE Board 2023 Exams
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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 30, 2023 by Teachoo
Ex 2.2, 6 Write the function in the simplest form: tanβ1 1/β(π₯^2β1), |x| > 1 tanβ1 (1/β(π₯^2 β 1)) Putting x = sec ΞΈ = tanβ1 (1/β(γπππγ^πβ‘π½ β 1)) = tanβ1 (1/β(γ(π + γπππγ^πγβ‘π½ ) β 1)) = tanβ1 (1/β(tan^2β‘ΞΈ )) = tanβ1 (1/tanβ‘ΞΈ ) We write 1/β(π₯^2 β 1) in form of tan Whenever there is β(π₯^2β1) , we put x = sec ΞΈ = tanβ1 (cot ΞΈ) = tanβ1 tan (90 β ΞΈ) = 90 β ΞΈ = π /π β ΞΈ We assumed x = sec ΞΈ sec ΞΈ = x ΞΈ = sec-1 x Hence, our equation becomes tan-1 (1/β(π₯^2β1)) = π/2 β ΞΈ = π /π β secβ1 x (cot ΞΈ = tan (90 β ΞΈ) )