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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

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Ex 2.2, 11 Deleted for CBSE Board 2023 Exams You are here

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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 May 12, 2021 by Teachoo

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)