Question

If ${\cos }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=2k$, show that $y\frac{dy}{dx}=x{\tan }^{2}k$.

Answer 1

${\cos }^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=2k$

$\Rightarrow \left(\frac{x^2-y^2}{x^2+y^2}\right)=\cos 2k$

$\Rightarrow \left(\frac{x^2-y^2}{x^2+y^2}\right)=\frac{1-{\tan }^{2}k}{1+{\tan }^{2}k}$

$\Rightarrow \left(\frac{1-\frac{y^2}{x^2}}{1+\frac{y^2}{x^2}}\right)=\frac{1-{\tan }^{2}k}{1+{\tan }^{2}k}$

On comparing

$\frac{y^2}{x^2}={\tan }^{2}k$

$\Rightarrow y^2=x^2{\tan }^{2}k$

$\Rightarrow 2y\frac{dy}{dx}=2x{\tan }^{2}k$

$\Rightarrow y\frac{dy}{dx}=x{\tan }^{2}k$

Hence proved.