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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 2.2, 12 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.