Last updated at Aug. 11, 2021 by Teachoo

Transcript

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) )

Ex 2.2

Ex 2.2,1
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Ex 2.2, 2 Important Deleted for CBSE Board 2022 Exams

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Ex 2.2, 6 Deleted for CBSE Board 2022 Exams You are here

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Ex 2.2, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams

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Ex 2.2, 21 (MCQ) Deleted for CBSE Board 2022 Exams

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.