

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 You are here
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
Last updated at May 12, 2021 by Teachoo
Ex 2.2, 18 tan (sin−1 3/5 + cot−1 3/2 ) We write sin-1 3/5 & cot-1 3/2 in terms of tan-1 Let sin−1 3/5 = a & cot−1 3/2 = b So, our equation becomes tan (sin−1 𝟑/𝟓 + cot−1 𝟑/𝟐 ) = tan (a + b) = 𝒕𝒂𝒏〖𝒂 + 𝒕𝒂𝒏𝒃 〗/(𝟏 − 𝒕𝒂𝒏〖𝒂 𝒕𝒂𝒏𝒃 〗 ) Finding tan a Since a = sin−1 𝟑/𝟓 sin a = 3/5 cos a = √(1 −𝑠𝑖𝑛2𝑎) = √(1 −(3/5)^2 ) = √(16/25) = 4/5 tan a = sin𝑎/cos𝑎 = (3/5)/(4/5) = 3/5 × 5/4 = 𝟑/𝟒 Finding tan b Since b = cot −1 3/2 cot b = 3/2 tan b = 1/cot𝑏 = 1/(3/2) = 𝟐/𝟑 From (1) tan ("sin−1 " 𝟑/𝟓 " + cot−1 " 𝟑/𝟐) = tan〖𝑎 + tan𝑏 〗/(1 − tan〖𝑎 tan𝑏 〗 ) Putting tan a = 3/4 & tan b = 2/3 = (3/4 + 2/3 )/(1 − 3/4 × 2/3) = ((3(3) + 2(4) )/(4 × 3) )/( (4 × 3 − 3 × 2)/(4 × 3) ) = ((9 + 8 )/(4 × 3) )/( (12 − 6)/(4 × 3) ) = ((17 )/(4 × 3) )/( 6/(4 × 3) ) = 𝟏𝟕/𝟔