Ex 2.2, 8 - Chapter 2 Class 12 Inverse Trigonometric Functions

Last updated at April 16, 2024 by

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Ex 2.2, 8 Find the value of tan-1 ["2 cos " (2"sinβˆ’1" 1/2)] Solving sin-1 (𝟏/𝟐) Let y = sin-1 (1/2) sin y = 1/2 sin y = sin (𝝅/πŸ”) Range of principal value of sin βˆ’1 is [(βˆ’πœ‹)/2, ( πœ‹)/2] Hence, y = 𝝅/πŸ” Rough We know that sin 30Β° = 1/2 ΞΈ = 30Β° = 30 Γ— πœ‹/180 = πœ‹/6 Since 1/2 is positive Principal value is ΞΈ i.e. 𝝅/πŸ” Now solving tan-1 ["2 cos " (2"sinβˆ’1" 1/2)] = tan-1 ["2 cos " (2 Γ— πœ‹/6)] = tan-1 ["2 cos" 𝝅/πŸ‘] = tan-1 ["2" Γ—πŸ/𝟐] = tan-1 [1] = tan-1 [π­πšπ§β‘γ€–π…/πŸ’γ€— ] = 𝝅/πŸ’

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.