


Ex 2.2
Ex 2.2, 2 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 3 Deleted for CBSE Board 2022 Exams
Ex 2.2, 4 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 5 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 6 Deleted for CBSE Board 2022 Exams
Ex 2.2, 7 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 8 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 9 Deleted for CBSE Board 2022 Exams
Ex 2.2, 10 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 11 Deleted for CBSE Board 2022 Exams
Ex 2.2, 12 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 13 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 14 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 15 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 16 Deleted for CBSE Board 2022 Exams
Ex 2.2, 17 Deleted for CBSE Board 2022 Exams
Ex 2.2, 18 Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 20 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 2.2, 21 (MCQ) Deleted for CBSE Board 2022 Exams You are here
Last updated at Aug. 11, 2021 by Teachoo
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