If x-secθ-cosθ, y=secnθ-cosnθ, then dydx2

x=secθ-cosθ, y=secnθ-cosnθ dxdθ=secθtanθ+sinθ=tanθ(secθ+cosθ) dydθ=nsecnθtanθ+ncosn-1θsinθ=ntanθ(secnθ+cosnθ) dydx=ntanθ(secnθ+cosnθ)tanθ(secθ+cosθ)=n(secnθ+cosnθ)(secθ+cosθ) dydx2=n2(secnθ-cosnθ)2+4(secθ-cosθ)2+4=n2y2+4x2+4

If x-secθ-cosθ, y=secnθ-cosnθ, then dydx2

x=secθ-cosθ, y=secnθ-cosnθ dxdθ=secθtanθ+sinθ=tanθ(secθ+cosθ) dydθ=nsecnθtanθ+ncosn-1θsinθ=ntanθ(secnθ+cosnθ) dydx=ntanθ(secnθ+cosnθ)tanθ(secθ+cosθ)=n(secnθ+cosnθ)(secθ+cosθ) dydx2=n2(secnθ-cosnθ)2+4(secθ-cosθ)2+4=n2y2+4x2+4

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$If x - secθ - cosθ, y = \sec^{n} θ - \cos^{n} θ, then {(\frac{dy}{dx})}^{2}$

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$x^{2} (x^{2} + 4)$

$\frac{x^{2} + 4}{y^{2} + 4}$

$\frac{n^{2} (y^{2} + 4)}{x^{2} + 4}$

$\frac{y^{2} + 4}{x^{2} + 4}$

answer is C.

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Detailed Solution

$x = secθ - cosθ, y = \sec^{n} θ - \cos^{n} θ \frac{dx}{dθ} = secθtanθ + sinθ = tanθ (secθ + cosθ) \frac{dy}{dθ} = {nsec}^{n} θtanθ + {ncos}^{n - 1} θsinθ = ntanθ (\sec^{n} θ + \cos^{n} θ) \frac{dy}{dx} = \frac{ntanθ (\sec^{n} θ + \cos^{n} θ)}{tanθ (secθ + cosθ)} = \frac{n (\sec^{n} θ + \cos^{n} θ)}{(secθ + cosθ)} {(\frac{dy}{dx})}^{2} = \frac{n^{2} {(\sec^{n} θ - \cos^{n} θ)}^{2} + 4}{{(secθ - cosθ)}^{2} + 4} = n^{2} (\frac{y^{2} + 4}{x^{2} + 4})$