If x=a(cost+logtant2, y=a sint then dydx=

A
tant
B
-tant
C
$\tan (\frac{3 t}{2})$
D
$- \cot (\frac{3 t}{2})$

Solution:

Given $x = a (cost + logtan \frac{t}{2}), y = a sint$

$\begin{array}{rcl} \frac{d x}{d t} & = & \frac{d}{d t} (a \cos t + a \log \tan (\frac{t}{2})) \\ = & - a \sin t + a \frac{s e c^{2} (\frac{t}{2})}{2 \tan (\frac{t}{2})} \\ = & - a \sin t + a \frac{1}{\sin t} \\ = & \frac{a (- \sin^{2} t + 1)}{\sin t} \\ = & a \frac{\cos^{2} t}{\sin t} \end{array}$

$\frac{d y}{d t} = a \cos t$

Hence,

$\begin{array}{rcl} \frac{d y}{d x} & = & \frac{d y}{d t} \cdot \frac{d t}{d x} \\ = & a \cos t \cdot \frac{\sin t}{a \cos^{2} t} \\ = & \frac{\sin t}{\cos t} \\ = & \tan t \end{array}$

If $x = a ((cost + logtan \frac{t}{2}), y = a sint$ then $\frac{dy}{dx} =$

Solution:

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