A) \[\frac{\sin \,2y}{\sin \,2x}\]
B) \[-\frac{\sin \,2x}{\sin \,2y}\]
C) \[-\frac{\sin \,2y}{\sin \,2x}\]
D) \[\frac{\sin \,2y}{\sin \,2x}\]
Correct Answer: D
Solution :
\[{{\sin }^{2}}x+{{\cos }^{2}}y=1\] \[\Rightarrow 2\sin \,x\,cos\,x+2cos\,y\left( -\operatorname{siny}\frac{dy}{dx} \right)=0\] \[\Rightarrow \frac{dy}{dx}=\frac{-\sin 2x}{-\sin 2y}=\frac{\sin 2x}{\sin 2y}\]You need to login to perform this action.
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