If x = a(t - sin t) and y = a(1 - cos t), the | vedclass

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(C) Given $x = a(t - \sin t)$ and $y = a(1 - \cos t).$ First,differentiate $x$ with respect to $t$: $\frac{dx}{dt} = a(1 - \cos t).$ Next,differentiat

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$\cot \left( \frac{t}{2} \right)$

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(C) Given $x = a(t - \sin t)$ and $y = a(1 - \cos t).$ First,differentiate $x$ with respect to $t$: $\frac{dx}{dt} = a(1 - \cos t).$ Next,differentiate $y$ with respect to $t$: $\frac{dy}{dt} = a(0 - (-\sin t)) = a \sin t.$ Now,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1 - \cos t)} = \frac{\sin t}{1 - \cos t}.$ Using trigonometric identities $\sin t = 2 \sin \left( \frac{t}{2} \right) \cos \left( \frac{t}{2} \right)$ and $1 - \cos t = 2 \sin^2 \left( \frac{t}{2} \right):$ $\frac{dy}{dx} = \frac{2 \sin (t/2) \cos (t/2)}{2 \sin^2 (t/2)} = \frac{\cos (t/2)}{\sin (t/2)} = \cot \left( \frac{t}{2} \right).$

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

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