A) \[\cot \theta \]
B) \[\cos \theta \]
C) \[\tan \theta \]
D) \[\sin \theta \]
Correct Answer: C
Solution :
\[x=2\left[ \cos \theta +\log \left( \tan \frac{\theta }{2} \right) \right],\,\,y=2\sin \theta \] \[\frac{dx}{d\theta }=2\left[ -\sin \theta +\frac{1}{\tan \frac{\theta }{2}}.{{\sec }^{2}}\frac{\theta }{2}.\frac{1}{2} \right],\] \[\frac{dy}{d\theta }=2\,\cos \,\theta \] \[\Rightarrow \] \[\frac{dx}{d\theta }=2\left[ -\sin \theta +\frac{1}{2\sin \frac{\theta }{2}\,cos\,\frac{\theta }{2}} \right]\] \[\Rightarrow \] \[\frac{dx}{d\theta }=2\left[ -\sin \theta +\frac{1}{\,\sin \frac{\theta }{2}} \right]\] \[\Rightarrow \] \[\frac{dx}{d\theta }=2\left[ \frac{1-{{\sin }^{2}}\theta }{\sin \theta } \right]\] \[=\frac{2\,{{\cos }^{2}}\theta }{\sin \theta }\] Now, \[\frac{dy}{dx}=\frac{dy}{d\theta }/\frac{dx}{d\theta }=\frac{2\,\cos \,\theta }{2\,{{\cos }^{2}}\theta }\,\sin \theta \] \[=\frac{\sin \theta }{\cos \,\theta }=\tan \theta \]
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