MathematicsIf  x=secθ−cosθ, y=secnθ−cosnθ, then  (x2+4)(dydx)2=

If  x=secθcosθ,y=secnθcosnθ, then  (x2+4)(dydx)2=

  1. A

    n2(y24)

  2. B

    n2(4y2)

  3. C

    n2(y2+4)

  4. D

    n2(y+4)

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    Solution:

    x=secθcosθ

    dxdθ=secθtanθ+sinθ

    =secθtanθ+cosθtanθ

    =tanθ(secθ+tanθ)

    y=secnθcosnθ

    dydθ=nsecn1θsecθtanθncosn1θ(sinθ)

    =nsecnθtanθ+ncosn1θsinθ

    =nsecnθtanθ+ncosnθtanθ

    =ntanθ(secnθ+cosnθ)

    dydx=dydθdxdθ=n(secnθ+cosnθ)secθ+cosθ

    dydx=n(secnθ+cosnθ)secθ+cosθ

    (dydx)2=n2(secnθ+cosnθ)2(secθ+cosθ)2

    x2+4=(sec2θ+cos2θ2)+4

    =sec2θ+cos2θ+2

    =(secθ+cosθ)2

    (x2+4)(dydx)2=n2(secnθ+cosnθ)2(secθ+cosθ)2×(secθ+cosθ)2

    =n2(secnθ+cosnθ)2

    =n2(sec2nθ+cos2nθ+2)

    =n2(sec2nθ+cos2nθ+2+22)

    =n2((secnθcosnθ)2+4)

    =n2(y2+4)

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