Question

If $\sqrt{1-x^{2n}}+\sqrt{1-y^{2n}}=a\left(x^{n}-y^{n}\right)$, then $\sqrt{\frac{1-x^{2n}}{1-y^{2n}}} \frac{dy}{dx}$ is equal to

• A. $1$
• B. $x/y$
• C. $\frac{x^{n-1}}{y^{n-1}}$
• D. None of these

Answer 1

Answer: (C)

Put $x^{n}=cos\,\alpha, y^{n}=cos\,\beta$
$\Rightarrow a=\frac{sin\,\alpha+sin\,\beta}{cos\,\alpha-cos\,\beta}=\frac{2\,sin\left(\frac{\alpha+\beta}{2}\right)cos\left(\frac{\alpha-\beta}{2}\right)}{-2\,sin\left(\frac{\alpha +\beta }{2}\right)sin\left(\frac{\alpha -\beta }{2}\right)}$
$=-cot\left(\frac{\alpha -\beta }{2}\right)$
$\Rightarrow 2\,cot^{-1}\left(-a\right)=\alpha-\beta$
$\Rightarrow cos^{-1}\left(x^{n}\right)-cos^{-1}\left(y^{n}\right)=2\,cot^{-1}\left(-a\right)$
$\Rightarrow \frac{y^{n-1}}{\sqrt{1-y^{2n}}} \frac{dy}{dx}=\frac{x^{n-1}}{\sqrt{1-x^{2n}}} \Rightarrow \sqrt{\frac{1-x^{2x}}{1-y^{2n}} \frac{dy}{dx}}=\frac{x^{n-1}}{y^{n-1}}$