Question

If $\sec \left(\frac{x+y}{x-y}\right)=a^2$, show that $\frac{dy}{dx}=\frac{y}{x}$

Answer 1

$\sec \left(\frac{x+y}{x-y}\right)=a^2$ or $\frac{x+y}{x-y}={\sec }^{-1}\left(a^2\right)$

differentiating both sides -

$\Rightarrow (x-y)(1+\frac{dy}{dx})-(x+y)(1-\frac{dy}{dx})=0$

$\Rightarrow x+x\frac{dy}{dx}-y-y\frac{dy}{dx}-x+x\frac{dy}{dx}-y+y\frac{dy}{dx}=0$

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

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

Hence, proved.