Question

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

Answer 1

Let us consider,

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

Differentiate w.r.t. $x$

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

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

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

$2x^3-2x^{2}y\frac{dy}{dx}+2x{y}^2-2y^3\frac{dy}{dx}-\left(2x^3+2x^{2}y\frac{dy}{dx}-2x{y}^2-2y^3\frac{dy}{dx}\right)=0$

$2x^3-2x^{2}y\frac{dy}{dx}+2x{y}^2-2y^3\frac{dy}{dx}-2x^3-2x^{2}y\frac{dy}{dx}+2x{y}^2+2y^3\frac{dy}{dx}=0$

$-4x^{2}y\frac{dy}{dx}+4x{y}^2=0$

$\frac{dy}{dx}=\frac{-4x{y}^2}{-4x^{2}y}$

$\frac{dy}{dx}=\frac{4x{y}^2}{4x^{2}y}$

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