Solve this

We are given with an equation $x \sin (a+y)+\sin a \cos (a+y)=0$, we have to prove that $\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$ by

using the given equation we will first find the value of $\frac{d y}{d x}$ 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,

$\tan (a+y)=\frac{-\sin a}{x}$

$\sec ^{2}(a+y) \frac{d y}{d x}=\frac{\sin a}{x^{2}}$

we can further solve it by using the given equation,

$\sec ^{2}(a+y) \frac{d y}{d x}=\frac{\tan ^{2}(a+y)}{\sin ^{2} a} \sin a$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sin ^{2}(\mathrm{a}+\mathrm{y})}{\sin \mathrm{a}}$