If x = a(cos t + log tan frac{t}{2}) and y = | vedclass

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(A) Given $x = a(\cos t + \log 	an \frac{t}{2})$ and $y = a \sin t$. Differentiating $y$ with respect to $t$: $\frac{dy}{dt} = a \cos t$  .....$(i)$

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$	an t$

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(A) Given $x = a(\cos t + \log 	an \frac{t}{2})$ and $y = a \sin t$. Differentiating $y$ with respect to $t$: $\frac{dy}{dt} = a \cos t$  .....$(i)$ Differentiating $x$ with respect to $t$: $\frac{dx}{dt} = a [-\sin t + \frac{1}{	an(t/2)} \cdot \sec^2(t/2) \cdot \frac{1}{2}]$ $= a [-\sin t + \frac{\cos(t/2)}{\sin(t/2)} \cdot \frac{1}{\cos^2(t/2)} \cdot \frac{1}{2}]$ $= a [-\sin t + \frac{1}{2 \sin(t/2) \cos(t/2)}]$ $= a [-\sin t + \frac{1}{\sin t}] = a [\frac{1 - \sin^2 t}{\sin t}] = a \frac{\cos^2 t}{\sin t}$  .....$(ii)$ Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t}{a \cos^2 t / \sin t} = \frac{\sin t}{\cos t} = 	an t$.

If $x = a(\cos t + \log \tan \frac{t}{2})$ and $y = a \sin t$,then $\frac{dy}{dx} = $

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