Last updated at May 29, 2018 by Teachoo

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 = β(1 βπ ππ2π) = β(1 β(3/5)^2 ) = β(1 β9/25) = β((25 β 9)/25) = β(16/25) = 4/5 tan a = sinβ‘π/cosβ‘π = (3/5)/(4/5) = 3/5 Γ 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 = β(1 βπ ππ2π) = β(1 β(3/5)^2 ) = β(1 β9/25) = β((25 β 9)/25) = β(16/25) = 4/5 tan a = sinβ‘π/cosβ‘π = (3/5)/(4/5) = 3/5 Γ 5/4 = 3/4 tan a = 3/4 a = tan-1 3/4

Chapter 2 Class 12 Inverse Trigonometric Functions

Serial order wise

About the Author

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.