# Solve this

Question:

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

Solution:

We are given with an equation $\tan ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=a$, we have to prove that $\frac{d y}{d x}=\frac{x}{y} \frac{(1-\tan a)}{(1+\tan a)}$ 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,

$\frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\tan a$

$x^{2}-y^{2}=\left(x^{2}+y^{2}\right) \tan a$

Now differentiating with respect to $x$, we get,

$2 x-2 y \frac{d y}{d x}=\left(2 x+2 y \frac{d y}{d x}\right) \tan a$

$\frac{d y}{d x}[y \tan a+y]=x-x \tan x$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}-\mathrm{x} \tan \mathrm{a}}{\mathrm{y}+\mathrm{y} \tan \mathrm{a}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{\mathrm{y}} \frac{(1-\tan \mathrm{a})}{(1+\tan \mathrm{a})}$