# Example 6 - Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

Last updated at May 12, 2021 by Teachoo

Last updated at May 12, 2021 by Teachoo

Transcript

Example 6 Write cotβ1 (1/β(π₯^2 β 1)), |π₯| > 1 in the simplest form. cot-1 (1/β(π₯^2 β 1)) Putting x = sec ΞΈ = cotβ1 (1/β(γπ¬ππγ^πβ‘π β 1)) = cotβ1 (1/β(γ(π + γπππ§γ^πγβ‘π½ ) β 1)) = cotβ1 (1/β(tan^2β‘ΞΈ )) We write 1/β(π₯^2 β 1) in form of cot Whenever there is β(π₯^2β1) , we put x = sec ΞΈ = cotβ1 (1/tanβ‘ΞΈ ) = cotβ1 (cot ΞΈ) = ΞΈ We assumed x = sec ΞΈ sec ΞΈ = x ΞΈ = secβ1 x Hence, our equation becomes cotβ1 (1/β(π₯^2β1)) = ΞΈ cotβ1 (1/β(π₯^2β1)) = secβ1 x

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Example 3 Important

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Example 5 Important Deleted for CBSE Board 2022 Exams

Example 6 Important Deleted for CBSE Board 2022 Exams You are here

Example 7 Deleted for CBSE Board 2022 Exams

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Example 9 Important

Example 10 Important Deleted for CBSE Board 2022 Exams

Example 11 Important Deleted for CBSE Board 2022 Exams

Example 12 Important Deleted for CBSE Board 2022 Exams

Example 13 Important Deleted for CBSE Board 2022 Exams

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

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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.