Ex 2.2, 21 - Chapter 2 Class 12 Inverse Trigonometric Functions
Last updated at Sept. 24, 2018 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 ∴ tan-1 √3 = π/3 Finding cot-1 (−√𝟑) Let x = cot-1 (−√3) cot x = – √3 cot x = cot 5π/6 Range of principal value of cot−1 is (0,π) Hence, principal value is 5π/6 ∴ 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.