# Ex 2.2,1 - 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

Ex 2.2, 1 3sin-1 π₯ = sin-1 (3π₯ β 4π₯^3), π₯β [β1/2,1/2] Solving R.H.S sin^(β1) (3π₯ β 4π₯^3) Putting x = sin π = sin^(β1) (3 sin π β 4 γ"sin" γ^3π) = sin^(β1) (sin 3π ) = 3π = 3 γπ ππγ^(β1) x (sin 3x = 3 sin x β 4 sin^3x) Now, x = sin π sinβ1 x = π = L.H.S Hence, proved.

Ex 2.2

Ex 2.2,1
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Ex 2.2, 4 Important Deleted for CBSE Board 2022 Exams

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Ex 2.2, 10 Important Deleted for CBSE Board 2022 Exams

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Ex 2.2, 12 Important Deleted for CBSE Board 2022 Exams

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Ex 2.2, 17 Deleted for CBSE Board 2022 Exams

Ex 2.2, 18 Important Deleted for CBSE Board 2022 Exams

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

Ex 2.2, 20 (MCQ) Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 21 (MCQ) 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.