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Ex 2.2
Ex 2.2, 2 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 3 Deleted for CBSE Board 2023 Exams
Ex 2.2, 4 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 5 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 6 Deleted for CBSE Board 2023 Exams
Ex 2.2, 7 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 8 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 9 Deleted for CBSE Board 2023 Exams
Ex 2.2, 10 Important Deleted for CBSE Board 2023 Exams
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Ex 2.2, 12 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 13 Important Deleted for CBSE Board 2023 Exams
Ex 2.2, 14 Important Deleted for CBSE Board 2023 Exams
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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
Last updated at March 22, 2023 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)