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If $x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$, then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$\tan t$
We have,
$$
x=a\left(\cos t+\log \tan \frac{t}{2}\right)
$$
Now, $\frac{d x}{d t}=a\left\{-\sin t+\frac{1}{\tan \frac{t}{2}} \cdot \sec ^{2} \frac{t}{2} \cdot \frac{1}{2}\right\}$
$$
=a\left\{-\sin t+\frac{1}{\sin t}\right\}
$$
$$
=a\left(\frac{1-\sin ^{2} t}{\sin t}\right)=\frac{a \cos ^{2} t}{\sin t}
$$
and $y=a \sin t$
$$
\frac{d y}{d t}=a \cos t
$$
$\mathrm{So}$,
$$
\begin{aligned}
\frac{d y}{d x} &=\frac{d y / d t}{d x / d t} \\
&=\frac{a(\cos t \sin t)}{a \cos ^{2} t}=\tan t
\end{aligned}
$$
$$
x=a\left(\cos t+\log \tan \frac{t}{2}\right)
$$
Now, $\frac{d x}{d t}=a\left\{-\sin t+\frac{1}{\tan \frac{t}{2}} \cdot \sec ^{2} \frac{t}{2} \cdot \frac{1}{2}\right\}$
$$
=a\left\{-\sin t+\frac{1}{\sin t}\right\}
$$
$$
=a\left(\frac{1-\sin ^{2} t}{\sin t}\right)=\frac{a \cos ^{2} t}{\sin t}
$$
and $y=a \sin t$
$$
\frac{d y}{d t}=a \cos t
$$
$\mathrm{So}$,
$$
\begin{aligned}
\frac{d y}{d x} &=\frac{d y / d t}{d x / d t} \\
&=\frac{a(\cos t \sin t)}{a \cos ^{2} t}=\tan t
\end{aligned}
$$
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