Question

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

Answer 1

$x=\frac{\sin y}{\sin (a+y)}\Rightarrow \frac{dx}{dy}=\frac{\sin (a+y)\cos y-\sin y\cos (a+y)}{{\sin }^2(a+y)}\Rightarrow \frac{dx}{dy}=\frac{\sin \left\{\right(a+y)-y\}}{{\sin }^2(a+y)}=\frac{\sin a}{{\sin }^2(a+y)}$

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