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(A) Given equations are $x=a(	heta-\sin 	heta)$ and $y=a(1+\cos 	heta)$.Differentiating $x$ with respect to $	heta$:$\frac{dx}{d	heta} = a[\frac{

Vedclass · Accepted Answer

$-\cot \frac{	heta}{2}$

Vedclass · Answer

(A) Given equations are $x=a(	heta-\sin 	heta)$ and $y=a(1+\cos 	heta)$.Differentiating $x$ with respect to $	heta$:$\frac{dx}{d	heta} = a[\frac{d}{d	heta}(	heta) - \frac{d}{d	heta}(\sin 	heta)] = a(1-\cos 	heta)$.Differentiating $y$ with respect to $	heta$:$\frac{dy}{d	heta} = a[\frac{d}{d	heta}(1) + \frac{d}{d	heta}(\cos 	heta)] = a(0 - \sin 	heta) = -a \sin 	heta$.Using the chain rule for parametric differentiation:$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{-a \sin 	heta}{a(1-\cos 	heta)}$.Using trigonometric identities $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ and $1-\cos 	heta = 2 \sin^2 \frac{	heta}{2}$:$\frac{dy}{dx} = \frac{-2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{2 \sin^2 \frac{	heta}{2}} = -\frac{\cos \frac{	heta}{2}}{\sin \frac{	heta}{2}} = -\cot \frac{	heta}{2}$.

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=a(\theta-\sin \theta)$ and $y=a(1+\cos \theta)$.

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