Question

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

Answer 1

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

$\frac{x^2-y^2}{x^2+y^2}=\cos \left\{{\tan }^{-1}a\right\}=$ constant

Differentiate w.r.t. $x$

$\Rightarrow \frac{d}{dx}\left(\frac{x^2-y^2}{x^2+y^2}\right)=0$

$\Rightarrow \frac{(x^2+y^2).\frac{d}{dx}(x^2-y^2)-(x^2-y^2).\frac{d}{dx}(x^2+y^2)}{(x^2+y^2{)}^2}$

$\Rightarrow (x^2+y^2)[2x-2y\frac{dy}{dx}]-(x^2-y^2)(2x+2y\frac{dy}{dx})=0$

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

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

Hence, $\frac{dy}{dx}=\frac{y}{x}$.