Chapter 2 Class 12 Inverse Trigonometric Functions

Class 12
Important Questions for exams Class 12

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Transcript

Ex 2.1, 14 (Method 1) Find the value of tan−1 √3 – sec−1(–2) is equal to π (B) – π/3 (C) π/3 (D) 2π/3 Solving tan−1 √𝟑 Let y = tan−1 √3 tan y = √3 tan y = tan (𝝅/𝟑) ∴ y = 𝝅/𝟑 Since Range of tan−1 is ((−𝜋)/2,𝜋/2) Hence, principal value is π/3 Solving sec−1 (–2) Let y = sec−1 (–2) y = 𝜋 − sec−1 (2) y = 𝜋 − 𝝅/𝟑 y = 𝟐𝝅/𝟑 Since range of sec−1 is [0,π] – {π/2} Hence, Principal Value is 𝟐𝛑/𝟑 We know that cos−1 (−x) = 𝜋 − cos −1 x Since sec a𝜋/3 = 2 𝜋/3 = sec−1 2 Solving sec−1 (–2) Let y = sec−1 (–2) y = 𝜋 − sec−1 (2) y = 𝜋 − 𝝅/𝟑 y = 𝟐𝝅/𝟑 Since range of sec−1 is [0,π] – {π/2} Hence, Principal Value is 𝟐𝛑/𝟑 We know that cos−1 (−x) = 𝜋 − cos −1 x Since sec a𝜋/3 = 2 𝜋/3 = sec−1 2 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) Now we have tan−1 (√3) = π/3 & sec−1 (–2) = 2π/3 Solving tan−1 √𝟑 – sec−1 (−2) = π/3 − (2π/3) = (π − 2π)/3 = (−𝛑)/𝟑 Hence , correct answer is (B) Ex 2.1, 14 (Method 2) Find the value of tan−1 √3 – sec−1(–2) is equal to π (B) – π/3 (C) π/3 (D) 2π/3 Solving tan−1 √𝟑 Let y = tan−1 √3 tan y = √3 tan y = tan (𝝅/𝟑) ∴ y = 𝝅/𝟑 Since Range of tan−1 is ((−𝜋)/2,𝜋/2) Hence, principal value is π/3 Solving sec−1 (–2) Let y = sec−1 (–2) sec y = −2 sec y = sec (𝟐𝝅/𝟑) Since range of sec−1 is [0,π] – {π/2} Hence, Principal Value is 2π/3 Now we have tan−1 (√3) = π/3 & sec−1 (–2) = 2π/3 Solving tan−1 √𝟑 – sec−1 (−2) = π/3 − (2π/3) = (π − 2π)/3 = (−𝛑)/𝟑 Hence , correct answer is (B)

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.