Question
If √(1−x2n)+√(1−y2n)=a(xn−yn), then √(1−x2n1−y2n)dydx is equal to
Answer: Option A
:
A
Putxn=sinθandyn=sinϕthen,(cosθ+cosϕ)=a(sinθ−sinϕ)⇒2cos(θ+ϕ2)cos(θ−ϕ2)=2acos(θ+ϕ2)sin(θ−ϕ2)⇒cot(θ−ϕ2)=a⇒(θ−ϕ2)=cot−1a⇒θ−ϕ=2cot−1aorsin−1xn−sin−1yn=2cot−1aDifferentiatingbothsides,wehavenxn−1√(1−x2n)−nyn−1√(1−y2n)dydx=0∴√(1−x2n1−y2n)dydx=xn−1yn−1
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:
A
Putxn=sinθandyn=sinϕthen,(cosθ+cosϕ)=a(sinθ−sinϕ)⇒2cos(θ+ϕ2)cos(θ−ϕ2)=2acos(θ+ϕ2)sin(θ−ϕ2)⇒cot(θ−ϕ2)=a⇒(θ−ϕ2)=cot−1a⇒θ−ϕ=2cot−1aorsin−1xn−sin−1yn=2cot−1aDifferentiatingbothsides,wehavenxn−1√(1−x2n)−nyn−1√(1−y2n)dydx=0∴√(1−x2n1−y2n)dydx=xn−1yn−1
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