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

Chapter 2 Class 12 Inverse Trigonometric Functions

Ex 2.1, 5
Important

Ex 2.1, 8 Important

Ex 2.1, 12 Important

Ex 2.1, 14 Important

Example 5 Important

Example 8 Important

Ex 2.2, 12 Important

Ex 2.2, 15 Important

Ex 2.2, 19 Important

Ex 2.2, 21 Important You are here

Example 10 Important

Example 12 Important

Example 13 Important

Misc. 2 Important

Misc. 7 Important

Misc. 10 Important

Misc. 11 Important

Misc 12 Important

Misc. 17 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.