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(A) Given equations are $x=a\left(\cos t+\log 	an \frac{t}{2}ight)$ and $y=a \sin t$.First,differentiate $x$ with respect to $t$:$\frac{dx}{dt} = a

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$	an t$

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(A) Given equations are $x=a\left(\cos t+\log 	an \frac{t}{2}ight)$ and $y=a \sin t$.First,differentiate $x$ with respect to $t$:$\frac{dx}{dt} = a \left[ \frac{d}{dt}(\cos t) + \frac{d}{dt}(\log 	an \frac{t}{2}) ight]$$= a \left[ -\sin t + \frac{1}{	an \frac{t}{2}} \cdot \sec^2 \frac{t}{2} \cdot \frac{1}{2} ight]$$= a \left[ -\sin t + \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} \cdot \frac{1}{\cos^2 \frac{t}{2}} \cdot \frac{1}{2} ight]$$= a \left[ -\sin t + \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}} ight]$$= a \left[ -\sin t + \frac{1}{\sin t} ight] = a \left( \frac{1 - \sin^2 t}{\sin t} ight) = a \frac{\cos^2 t}{\sin t}$.Next,differentiate $y$ with respect to $t$:$\frac{dy}{dt} = a \frac{d}{dt}(\sin t) = a \cos t$.Now,find $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t}{a \frac{\cos^2 t}{\sin t}} = \frac{\cos t \cdot \sin t}{\cos^2 t} = \frac{\sin t}{\cos t} = 	an t$.

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$.

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