A) \[\frac{y}{x}\]
B) \[\frac{x}{y}\]
C) 0
D) none of these
Correct Answer: D
Solution :
Given that,\[y=\cos t,x=\sin t\] On differentiating both sides w.r.t. t, we get \[\frac{dy}{dx}=-\sin \,t,\frac{dx}{dt}=\cos t\] \[\therefore \] \[\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}\] \[=\frac{-\sin t}{\cos t}=-\tan t\] Again differentiating, we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-{{\sec }^{2}}t.\frac{dt}{dx}\] \[=-{{\sec }^{2}}t.\frac{1}{\cos t}\] \[=-\frac{1}{{{\cos }^{3}}t}\] \[=-\frac{1}{{{y}^{3}}}\]
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