1. Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
  2. Serial order wise
  3. Ex 2.2

Ex 2.2, 11 - Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

Last updated at May 12, 2021 by

Ex 2.2, 11 - Find tan-1 [2 cos (2 sin-1 1/2)] - Class 12

Ex 2.2, 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2

  1. Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
  2. Serial order wise

    3. Ex 2.2

    Examples →

Transcript

Ex 2.2, 11 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’ (Since cosπœ‹/3 = 1/2)

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
Serial order wise

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