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

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

$2{\tan }^{-1}\frac{y}{x}=2k$(using inverse trigonometric formula)

$\therefore \tan k=\frac{y}{x}.....\left(1\right)$

$\frac{x^2-y^2}{x^2+y^2}=\cos 2k$

differentiate w.r.t. x

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

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

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

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

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

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

$\frac{dy}{dx}=\frac{y}{x}$........$\left(2\right)$

LHS of given equation $y\frac{dy}{dx}=y\times \frac{y}{x}=\frac{y^2}{x}$(from $2$)

RHS of given equation $x{\tan }^{2}k=x\times (\frac{y}{x}{)}^2=\frac{y^2}{x}$(from $1$)

$\therefore LHS=RHS$