A) \[\frac{\sin a}{{{\sin }^{2}}(a+y)}\]
B) \[\frac{{{\sin }^{2}}\,(a+y)}{\sin \,\,a}\]
C) \[\sin \,a\,{{\sin }^{2}}(a+y)\]
D) \[\frac{{{\sin }^{2}}(a-y)}{\sin a}\]
Correct Answer: B
Solution :
Since, \[\sin y=x\sin (a+y)\] \[\Rightarrow \] \[x=\frac{\sin y}{\sin (a+y)}\] On differentiating w.r.t. y, we get \[\frac{dx}{dy}=\frac{\sin (a+y)\cos y-\sin y\cos (a+y)}{{{\sin }^{2}}(a+y)}\] \[\Rightarrow \frac{dx}{dy}=\frac{\sin (a+y-y)}{{{\sin }^{2}}(a+y)}\] \[\Rightarrow \] \[\frac{dx}{dy}=\frac{\sin a}{{{\sin }^{2}}(a+y)}\] \[\Rightarrow \frac{dy}{dx}=\frac{{{\sin }^{2}}(a+y)}{\sin a}\]
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