Chapter 2 Class 12 Inverse Trigonometric Functions

Class 12
Important Questions for exams Class 12

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### 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) ∴ y = 𝝅/𝟑 Since range of tan-1 is ((−π)/2,π/2) Hence, Principal Value is 𝝅/𝟑 (Since tan 𝜋/3 = √3) ∴ tan−1 √3 = π/3 Finding cot−1 (−√𝟑) Let x = cot−1 (√3) x = 𝜋 − cot−1 (√3) x = 𝜋 − 𝛑/𝟔 x = 𝟓𝛑/𝟔 We know that cot−1 (−x) = 𝜋 − cot −1 x Since cot 𝜋/6 = √3 𝜋/3 = cot−1 (√3) Since range 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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#### 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.