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) \[\frac{{{\sin }^{2}}(a+y)}{\cos a}\]
Correct Answer: C
Solution :
\[\sin y=x\sin (a+y)\]Þ\[x=\frac{\sin y}{\sin (a+y)}\] Þ \[1=\frac{\cos y.\frac{dy}{dx}.\sin (a+y)-\sin y\cos (a+y)\frac{dy}{dx}}{{{\sin }^{2}}(a+y)}\] \[=\frac{\frac{dy}{dx}.\sin (a+y-y)}{{{\sin }^{2}}(a+y)}\Rightarrow \frac{dy}{dx}=\frac{{{\sin }^{2}}(a+y)}{\sin a}\].
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