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Ex 2.2, 21 - Chapter 2 Class 12 Inverse Trigonometric Functions
Last updated at Feb. 13, 2020 by Teachoo
Transcript
Ex 2.2, 21 Find value of : tan−1 √3 – cot−1 (−√3) Finding tan−1 √𝟑 Let y = tan−1 √3 tan y = √3 tan y = tan (π/3) We know that principal value range of tan-1 is ((−π)/2,π/2) Hence, principal value of tan−1 √3 is π/3 Rough We know that tan 60° = √3 θ = 60° = 60 × 𝜋/180 = 𝜋/3 Since √3 is positive Principal value is θ i.e. 𝜋/3 ∴ cot−1 (−√3) = 5π/6 Hence, tan−1 √3 = π/3 & cot−1 (−√3) = 5π/6 Now calculating tan−1 √𝟑 – cot−1 (−√𝟑) = π/3 − 5π/6 = (2𝜋 − 5π)/6 = (−3π)/6 = (−𝝅)/𝟐 Hence, option (B) is correct
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Chapter 2 Class 12 Inverse Trigonometric Functions
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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.