Question

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

Answer 1

$\sin y=x\sin (a+y)$

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

Differentiating w.r.t. $x$ we get,
$\therefore 1=\frac{\sin (a+y)\cos y-\sin y\cos (a+y)}{{\sin }^2(a+y)}\frac{dy}{dx}$

$\therefore 1=\frac{\sin (a+y-y)}{{\sin }^2(a+y)}\frac{dy}{dx}$

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