If \[\sin y=x\sin (a+y),\]then \[\frac{dy}{dx}=\] [Karnataka CET 2000; UPSEAT 2001; Pb. CET 2001; Kerala (Engg.) 2005]

\[\frac{{{\sin }^{2}}(a+y)}{\cos a}\]

\[\frac{{{\sin }^{2}}(a+y)}{\sin (a+2y)}\]

\[\frac{{{\sin }^{2}}(a+y)}{\sin a}\]

\[\frac{{{\sin }^{2}}(a+y)}{\cos a}\]

If \[\sin y=x\sin (a+y),\]then \[\frac{dy}{dx}=\] [Karnataka CET 2000;

1.	If \[\sin y=x\sin (a+y),\]then \[\frac{dy}{dx}=\] [Karnataka CET 2000; UPSEAT 2001; Pb. CET 2001; Kerala (Engg.) 2005]
A.	\[\frac{{{\sin }^{2}}(a+y)}{\sin (a+2y)}\]
B.	\[\frac{{{\sin }^{2}}(a+y)}{\sin (a+2y)}\]
C.	\[\frac{{{\sin }^{2}}(a+y)}{\sin a}\]
D.	\[\frac{{{\sin }^{2}}(a+y)}{\cos a}\]
Answer» D. \[\frac{{{\sin }^{2}}(a+y)}{\cos a}\]