Last updated at May 12, 2021 by Teachoo

Transcript

Ex 2.2, 11 Find the value of tan-1 ["2 cos " (2"sinβ1" 1/2)] Solving sin-1 (π/π) Let y = sin-1 (1/2) sin y = 1/2 sin y = sin (π /π) Range of principal value of sin β1 is [(βπ)/2, ( π)/2] Hence, y = π /π Rough We know that sin 30Β° = 1/2 ΞΈ = 30Β° = 30 Γ π/180 = π/6 Since 1/2 is positive Principal value is ΞΈ i.e. π /π Now solving tan-1 ["2 cos " (2"sinβ1" 1/2)] = tan-1 ["2 cos " (2 Γ π/6)] = tan-1 ["2 cos" π /π] = tan-1 ["2" Γπ/π] = tan-1 [1] = tan-1 [πππ§β‘γπ /πγ ] = π /π (Since cosπ/3 = 1/2)

Ex 2.2

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Ex 2.2, 11 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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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.