Choose the correct alternative in the following:
If $\sin y=x \sin (a+y)$, then $\frac{d y}{d x}$ is
A. $\frac{\sin a}{\sin a \sin ^{2}(a+y)}$
B. $\frac{\sin ^{2}(a+y)}{\sin a}$
C. $\sin a \sin ^{2}(a+y)$
D. $\frac{\sin ^{2}(a-y)}{\sin a}$
$\sin y=x \sin (a+y)$
$\Rightarrow \frac{\sin y}{\sin (a+y)}=x$
Differentiating w.r.t y we get,
$\Rightarrow \frac{d x}{d y}=\frac{d}{d y}\left(\frac{\sin y}{\sin (a+y)}\right)$
$\Rightarrow \frac{d x}{d y}=\frac{d}{d y}\left(\frac{\sin y}{\sin (a+y)}\right)$
$=\frac{\cos y(\sin (a+y))-\cos (a+y) \cdot \sin y}{[\sin (a+y)]^{2}}$
$=\frac{\cos y(\sin a \cos y+\cos a \sin y)-(\cos a \cos y-\sin a \sin y) \sin y}{[\sin (a+y)]^{2}}$
$=\frac{\sin a \cos ^{2} y+\cos a \cos y \sin y-\sin y \cos a \cos y+\sin a \sin ^{2} y}{[\sin (a+y)]^{2}}$
$=\frac{\sin a\left(\cos ^{2} y+\sin ^{2} y\right)+\cos a \cos y \sin y-\sin y \cos a \cos y}{[\sin (a+y)]^{2}}$
$\frac{d x}{d y}=\frac{\sin a}{[\sin (a+y)]^{2}} \because \cos ^{2} y+\sin ^{2} y=1$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sin ^{2}(\mathrm{a}+\mathrm{y})}{\sin \mathrm{a}}=(\mathrm{B})$