1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
  3. Ex 2.2
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Ex 2.2, 8 - Chapter 2 Class 12 Inverse Trigonometric Functions

Last updated at Dec. 16, 2024 by Teachoo

Transcript

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’
  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise

    3. Ex 2.2

    Examples →
Examples →
Chapter 2 Class 12 Inverse Trigonometric Functions
Serial order wise

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Davneet Singh

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 and Computer Science at Teachoo