Ex 2.2, 8 - Chapter 2 Class 12 Inverse Trigonometric Functions
Last updated at Dec. 16, 2024 by Teachoo
Finding principal value
Principal Value
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Chapter 2 Class 12 Inverse Trigonometric Functions
Concept wise
Transcript
Ex 2.2, 8 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
