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Ex 2.2

Ex 2.2,1
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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

Ex 2.2, 6 Deleted for CBSE Board 2023 Exams You are here

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

Chapter 2 Class 12 Inverse Trigonometric Functions

Serial order wise

Last updated at Aug. 11, 2021 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) )