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Ex 2.2, 18 - tan (sin-1 3/5 + cot-1 3/2) - Class 12 NCERT

Ex 2.2, 18 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Ex 2.2, 18 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3

 

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Transcript

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 + 𝒕𝒂𝒏⁡𝒃 〗/(𝟏 − 𝒕𝒂𝒏⁡〖𝒂 𝒕𝒂𝒏⁡𝒃 〗 ) 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 = sin⁡𝑎/cos⁡𝑎 = (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/cot⁡𝑏 = 1/(3/2) = 𝟐/𝟑 From (1) tan ("sin−1 " 𝟑/𝟓 " + cot−1 " 𝟑/𝟐) = tan⁡〖𝑎 + tan⁡𝑏 〗/(1 − tan⁡〖𝑎 tan⁡𝑏 〗 ) 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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.