Question

If $\sqrt{1-x^{2n}}+\sqrt{1-y^{2n}}=a(x^n-y^n)$ then $\sqrt{\frac{1-x^{2n}}{1-y^{2n}}}.\frac{dy}{dx}=$

• A. $1$
• B. $\frac{x}{y}$
• C. $\frac{x^{n-1}}{y^{n-1}}$
• D. $\frac{y}{x}$

Answer 1

The correct option is C

$\frac{x^{n-1}}{y^{n-1}}$

Let $x^n=sinA;y^n=sinB$
After simplification, we get
$si{n}^{-1}\left(x^n\right)-si{n}^{-1}\left(y^n\right)=2co{t}^{-1}\left(a\right)$
diff. on both sides w.r.t.'x', we get
$\frac{n{x}^{n-1}}{\sqrt{1-x^{2n}}}-\frac{1}{\sqrt{1-y^{2n}}}.n{y}^{n-1}.\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{x^{n-1}}{y^{n-1}}.\sqrt{\frac{1-y^{2n}}{1-x^{2n}}}$