CBSE Class 12 Sample Paper for 2020 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## dy/dx = √(1 - y 2 )/√(1 - x 2 ).

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Question 28 (OR 1st Question) If √(1−𝑥^2 ) + √(1−𝑦^2 ) = a (x − y), then prove that 𝑑𝑦/(𝑑𝑥 ) = √(1 − 𝑦^2 )/√(1 − 𝑥^2 ). Finding 𝒅𝒚/𝒅𝒙 would be complicated here To make life easy, we substitute x = sin A y = sin B (As √(1−𝑥^2 )= √(1−sin^2⁡𝐴 )=√(cos^2⁡𝐴 )) And then solve Let’s substitute x = sin A y = sin B in our equation Now √(1−𝑥^2 ) + √(1−𝑦^2 ) = a (x − y) Putting x = sin A and y = sin B √(1−sin^2⁡𝐴 ) + √(1−sin^2⁡𝐵 ) = a (sin A − sin B) √(cos^2⁡𝐴 ) + √(cos^2⁡𝐵 ) = a (sin A − sin B) cos A + cos B = a (sin A − sin B) Using cos A + cos B = 2 cos (𝐴+𝐵)/2 cos (𝐴−𝐵)/2 and sin A – sin B = 2 cos (𝐴+𝐵)/2 sin (𝐴−𝐵)/2 2 cos⁡((𝐴 + 𝐵)/2) cos⁡((𝐴 − 𝐵)/2) = a × 2 cos⁡((𝐴 + 𝐵)/2) s𝑖𝑛⁡((𝐴 − 𝐵)/2) cos⁡((𝐴 − 𝐵)/2) = a s𝑖𝑛⁡((𝐴 − 𝐵)/2) 〖cos 〗⁡((𝐴 − 𝐵)/2)/〖sin 〗⁡((𝐴 − 𝐵)/2) = a cot⁡((𝐴 − 𝐵)/2) = a (𝐴 − 𝐵)/2 = 〖𝑐𝑜𝑡〗^(−1) 𝑎 𝐴−𝐵 = 2 〖𝑐𝑜𝑡〗^(−1) 𝑎 Putting back values of A and B sin^(−1)⁡𝑥−sin^(−1)⁡𝑦 = 2〖𝑐𝑜𝑡〗^(−1) 𝑎 Differentiating w.r.t x 1/√(1 − 𝑥^2 )−1/√(1 − 𝑦^2 )×𝑑𝑦/𝑑𝑥=0 1/√(1 − 𝑥^2 )=1/√(1 − 𝑦^2 )×𝑑𝑦/𝑑𝑥 √(1 − 𝑦^2 )/√(1 − 𝑥^2 )=𝑑𝑦/𝑑𝑥 𝑑𝑦/(𝑑𝑥 ) = √(1 − 𝑦^2 )/√(1 − 𝑥^2 ) Hence proved

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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.