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

Transcript

Ex 2.1, 11 Find the value of tan−1 (1) + cos−1 (−1/2) + sin-1 (−1/2) Solving tan−1 (1) Let y = tan−1 (1) tan y = 1 tan y = tan (π/4) Range of principal value of tan−1 is (−π/2,π/2) Hence, the principal value of tan-1 (1) is π/4 Rough We know that tan 45° = 1 θ = 45° = 45 × 𝜋/180 = 𝜋/4 Since 1 is positive Principal value is θ i.e. 𝜋/4 Solving cos−1 (−𝟏/𝟐) Let y = cos−1 (−1/2) cos y = −1/2 cos y = cos (2π/3) Range of principal value of cos−1 is [0 , 𝜋] Hence, the principal value is 2π/3. Rough We know that cos 60° = 1/2 θ = 60° = 60 × 𝜋/180 = 𝜋/3 Since (−1)/2 is negative Principal value is 𝜋 – θ i.e. π −𝜋/3 = 2𝜋/3 Solving sin−1 (−𝟏/𝟐) Let y = sin−1 (−1/2) sin y = (−1)/2 sin y = sin ((−π)/6) Range of principal value of sin−1 is between [(−𝜋)/2 " , " 𝜋/2] Hence, the principal value is (−π)/6 Rough We know that sin 30° = 1/2 θ = 30° = 30 × 𝜋/180 = 𝜋/6 Since (−1)/2 is negative Principal value is −θ i.e. (−𝜋)/6 Now we have tan−1 (1) = π/4 , cos−1 (−1/2) = 2π/3 , sin−1 (−1/2) = (−π)/6 Finding tan−1 (1) + cos−1 ((−𝟏)/𝟐) + sin−1 ((−𝟏)/𝟐) = π/4 + 2π/3 – π/6 = (3 × π + 4 × (2π) − 2 (π))/12 = (3π + 8π − 2π)/12 = 9π/12 = 𝟑𝛑/𝟒

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.