If √(1 - x ^{ 2 } ) + √(1 - y ^{ 2 } ) = a (x − y), then prove that
dy/dx = √(1 - y ^{ 2 } )/√(1 - x ^{ 2 } ).
CBSE Class 12 Sample Paper for 2020 Boards
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CBSE Class 12 Sample Paper for 2020 Boards
Last updated at April 16, 2024 by Teachoo
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^2𝐴 )=√(cos^2𝐴 )) 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^2𝐴 ) + √(1−sin^2𝐵 ) = a (sin A − sin B) √(cos^2𝐴 ) + √(cos^2𝐵 ) = 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