Ex 2.2, 18 - tan (sin-1 3/5 + cot-1 3/2) - Class 12 NCERT - Changing of trignometric variables and then applying formula

 

  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
Ask Download

Transcript

Ex 2.2, 18 (Method 1) tan (sin-1 3/5 + cot-1 3/2 ) Let sin-1 3/5 = a & cot-1 3/2 = b Now sin-1 3/5 = a 3/5 = sin a sin a = 3/5 cos a = √(1 −𝑠𝑖𝑛2𝑎) = √(1 −(3/5)^2 ) = √(1 −9/25) = √((25 − 9)/25) = √(16/25) = 4/5 tan a = sin⁡𝑎/cos⁡𝑎 = (3/5)/(4/5) = 3/5 × 5/4 = 3/4 Hence, tan (sin-1 3/5 + cot-1 3/2 ) = 17/6 Hence Proved Ex 2.2, 18 (Method 2) tan (sin-1 3/5 + cot-1 3/2 ) Let sin-1 3/5 = a & cot-1 3/2 = b Now sin-1 3/5 = a 3/5 = sin a sin a = 3/5 cos a = √(1 −𝑠𝑖𝑛2𝑎) = √(1 −(3/5)^2 ) = √(1 −9/25) = √((25 − 9)/25) = √(16/25) = 4/5 tan a = sin⁡𝑎/cos⁡𝑎 = (3/5)/(4/5) = 3/5 × 5/4 = 3/4 tan a = 3/4 a = tan-1 3/4

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail