If x=a(t+sin t) and y=a(1-cos t),then frac{dy | vedclass

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(A) Given $x=a(t+\sin t)$ and $y=a(1-\cos t)$.Differentiating $x$ with respect to $t$,we get $\frac{dx}{dt} = a(1+\cos t)$.Differentiating $y$ with re

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$	an \frac{t}{2}$

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(A) Given $x=a(t+\sin t)$ and $y=a(1-\cos t)$.Differentiating $x$ with respect to $t$,we get $\frac{dx}{dt} = a(1+\cos t)$.Differentiating $y$ with respect to $t$,we get $\frac{dy}{dt} = a(\sin t)$.Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1+\cos t)}$.Using trigonometric identities $\sin t = 2 \sin \frac{t}{2} \cos \frac{t}{2}$ and $1+\cos t = 2 \cos^2 \frac{t}{2}$,we get:$\frac{dy}{dx} = \frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos^2 \frac{t}{2}} = 	an \frac{t}{2}$.

If $x=a(t+\sin t)$ and $y=a(1-\cos t)$,then $\frac{dy}{dx}=$

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