If x=3left[sin t-log left(cot frac{t}{2}right | vedclass

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(C) Given,$x=3\left[\sin t-\log \left(\cot \frac{t}{2}ight)ight]$ and $y=6\left[\cos t+\log \left(	an \frac{t}{2}ight)ight]$.First,differenti

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$\frac{2 \cos^2 t}{1+\sin t \cos t}$

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(C) Given,$x=3\left[\sin t-\log \left(\cot \frac{t}{2}ight)ight]$ and $y=6\left[\cos t+\log \left(	an \frac{t}{2}ight)ight]$.First,differentiate $x$ with respect to $t$:$\frac{dx}{dt} = 3\left[\cos t - \frac{1}{\cot(t/2)} \cdot (-\csc^2(t/2)) \cdot \frac{1}{2}ight] = 3\left[\cos t + \frac{\csc^2(t/2)}{2 \cot(t/2)}ight] = 3\left[\cos t + \frac{1}{2 \sin(t/2) \cos(t/2)}ight] = 3\left[\cos t + \frac{1}{\sin t}ight] = \frac{3(1+\sin t \cos t)}{\sin t}$.Next,differentiate $y$ with respect to $t$:$\frac{dy}{dt} = 6\left[-\sin t + \frac{1}{	an(t/2)} \cdot \sec^2(t/2) \cdot \frac{1}{2}ight] = 6\left[-\sin t + \frac{\sec^2(t/2)}{2 	an(t/2)}ight] = 6\left[-\sin t + \frac{1}{2 \sin(t/2) \cos(t/2)}ight] = 6\left[-\sin t + \frac{1}{\sin t}ight] = 6\left(\frac{1-\sin^2 t}{\sin t}ight) = \frac{6 \cos^2 t}{\sin t}$.Finally,calculate $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:$\frac{dy}{dx} = \frac{6 \cos^2 t / \sin t}{3(1+\sin t \cos t) / \sin t} = \frac{2 \cos^2 t}{1+\sin t \cos t}$.

If $x=3\left[\sin t-\log \left(\cot \frac{t}{2}\right)\right]$ and $y=6\left[\cos t+\log \left(\tan \frac{t}{2}\right)\right]$,then $\frac{dy}{dx}=$

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