Question

If $x\sin (a+y)+\sin a\cos (a+y)=0$, show that $\frac{dy}{dx}=\frac{{\sin }^2(a+y)}{\sin a}$.

Answer 1

$x\sin (a+y)+\sin a\cos (a+y)=0\Rightarrow x\sin (a+y)=-\left(\sin a\cos (a+y)\right)\Rightarrow \frac{x}{\sin a}=-\frac{\cos (a+y)}{\sin (a+y)}\Rightarrow \frac{x}{\sin a}=-\cot (a+y)$

Differentiate the function

$\Rightarrow \frac{dx}{\sin a}=-(-cose{c}^2(a+y)dy\Rightarrow \frac{dy}{dx}=\frac{{\sin }^2(a+y)}{\sin a}$