Question

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

Answer 1

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

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

Let $\frac{y}{x}=\tan \theta$

$\Rightarrow {\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)={\tan }^{-1}\left(\alpha \right)$

$\Rightarrow {\cos }^{-1}\cos 2\theta ={\tan }^{-1}\left(\alpha \right)$

$\Rightarrow 2\theta ={\tan }^{-1}\alpha$

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

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

difference with respect to $x$ in both side -

$\Rightarrow \frac{x.\frac{dy}{dx}-y}{x^2}=0$

$\Rightarrow x\frac{dy}{dx}=y$

$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$

Hence, the answer is proved.