MathematicsIf x=cosec⁡θ−sin⁡θ,y=cosecn⁡θ−sinn⁡θthen(x2+4)(dydx)2−n2y2=

If x=cosecθsinθ,y=cosecnθsinnθthen(x2+4)(dydx)2n2y2=

  1. A

    n2

  2. B

    2n2

  3. C

    3n2

  4. D

    4n2

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    Solution:

     As x=cosecθsinθ, we have 

    x2+4=(cosecθsinθ)2+4=(cosecθ+sinθ)2

     and y2+4=cosecnθsinnθ2+4=cosecnθ+sinnθ2

    dydx=dydθdxdθ=ncosecn1θ(cosecθcotθ)nsinn1θcosθcosecθcotθcosθ=ncosecnθcotθ+sinn1θcosθ(cosecθcotθ+cosθ)=ncotθcosecnθ+sinnθcotθ(cosecθ+sinθ)=ncosecnθ+sinnθ(cosecθ+sinθ)=ny2+4x2+4 

     Squaring both sides, we get dydx2=n2y2+4x2+4

    x2+4dydx2=n2y2+4(x2+4)(dydx)2n2y2=4n2

     

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