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If $x=a(t+\sin t), y=a(1-\cos t)$, then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$\tan \frac{t}{2}$
$$
\begin{aligned}
& x=a(t+\sin t) \text { and } y=a(1-\cos t) \\
& \therefore \frac{d x}{d t}=a(1+\cos t) \text { and } \frac{d y}{d t}=a(\sin t) \\
& \therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a \sin t}{a(1+\cos t)}=\frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos ^2 \frac{t}{2}}=\tan \frac{t}{2}
\end{aligned}
$$
\begin{aligned}
& x=a(t+\sin t) \text { and } y=a(1-\cos t) \\
& \therefore \frac{d x}{d t}=a(1+\cos t) \text { and } \frac{d y}{d t}=a(\sin t) \\
& \therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a \sin t}{a(1+\cos t)}=\frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos ^2 \frac{t}{2}}=\tan \frac{t}{2}
\end{aligned}
$$
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