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Ex 2.1, 11 - Find value tan-1 (1) + cos-1 (-1/2) + sin-1 (-1/2)

Ex 2.1, 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Ex 2.1, 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3
Ex 2.1, 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 4
Ex 2.1, 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 5
Ex 2.1, 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 6
Ex 2.1, 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 7
Ex 2.1, 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 8

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Ex 2.1, 11 (Method 1) 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) ∴ y = 𝝅/𝟒 Since Range of tan−1 is (−π/2,π/2) Hence, the Principal Value is 𝝅/𝟒 Solving cos−1 ((−𝟏)/𝟐) Let y = cos−1 ((−1)/2) y = 𝜋 − cos−1 (1/2) y = 𝜋 − 𝝅/𝟑 y = 𝟐𝝅/𝟑 Since Range of cos−1 is [0 , 𝜋] Hence, the Principal Value is 2π/3 We know that cos−1 (−x) = 𝜋 − cos −1 x Since cos 𝜋/3 = 1/2 𝜋/3 = cos−1 (1/2) Solving sin−1 ((−𝟏)/𝟐) Let y = sin−1 ((−1)/2) y = − sin−1 (1/2) y = (−𝝅)/𝟔 Since Range of sin−1 is between [(−𝜋)/2 ", " 𝜋/2] Hence, Principal Value is (−𝝅)/𝟔 We know that sin−1 (−x) = − sin−1 x Since sin 𝜋/6 = 1/2 𝜋/6 = sin−1 (1/2) 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 Ex 2.1, 11 (Method 2) 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) ∴ y = 𝝅/𝟒 Since Range of tan−1 is (−π/2,π/2) Hence, the Principal Value is 𝝅/𝟒 Solving cos−1 ((−𝟏)/𝟐) Let y = cos−1 ((−1)/2) cos y = (−1)/2 cos y = cos (𝟐𝝅/𝟑) Since Range 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) Since Range of sin−1 is between [(−𝜋)/2 ", " 𝜋/2] Hence, 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.