If sin Y=X sin (a + y), then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to 

Solution:

$\because \sin \mathrm{y}=\mathrm{xsin}(\mathrm{a}+\mathrm{y})$

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

On differentiating  w.r.t.y, we get

$\begin{array}{l}\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\sin (\mathrm{a}+\mathrm{y})\cos \mathrm{y}-\sin \mathrm{ycos}(\mathrm{a}+\mathrm{y})}{{\sin }^2(\mathrm{a}+\mathrm{y})}\\ \Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\sin \mathrm{a}}{{\sin }^2(\mathrm{a}+\mathrm{y})}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\sin }^2(\mathrm{a}+\mathrm{y})}{\sin \mathrm{a}}\end{array}$