**Question:**

If $\sec \left(\frac{x+y}{x-y}\right)=a, p$ rove that $\frac{d y}{d x}=\frac{y}{x}$

**Solution:**

We are given with an equation $\sec \left(\frac{x+y}{x-y}\right)=a$, we have to prove that $\frac{d y}{d x}=\frac{y}{x}$ by using the given equation we

will first find the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to $x$, we get,

$\sec \left(\frac{x+y}{x-y}\right) \tan \left(\frac{x+y}{x-y}\right)\left[\frac{(x-y)\left(1+\frac{d y}{d x}\right)-(x+y)\left(1-\frac{d y}{d x}\right)}{(x-y)^{2}}\right]=0$

$\left[\frac{(x-y)\left(1+\frac{d y}{d x}\right)-(x+y)\left(1-\frac{d y}{d x}\right)}{(x-y)^{2}}\right]=0$

$-2 y+2 x \frac{d y}{d x}=0$

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