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The length of the subtangent at $t$ on the curve $\mathrm{x}=\mathrm{a}(\mathrm{t}+\sin \mathrm{t}), \mathrm{y}=\mathrm{a}(1-\cos \mathrm{t})$ is
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Verified Answer
The correct answer is:
$a \sin \mathrm{t}$
Given, $x=a(t+\sin t), y=a(1-\cos t)$
$$
\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a}(1+\cos \mathrm{t}), \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a}(\sin \mathrm{t})
$$
$$
\therefore \quad \frac{d y}{d x}=\frac{a \sin t}{a(1+\cos t)}=\tan \frac{t}{2}
$$
$$
\begin{aligned}
\therefore \text { Length of subtangent } &=\frac{y}{d y / d x} \\
&=\frac{a(1-\cos t)}{\tan \frac{t}{2}} \\
&=2 a \sin \frac{t}{2} \cos \frac{t}{2} \\
&=a \sin t
\end{aligned}
$$
$$
\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a}(1+\cos \mathrm{t}), \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a}(\sin \mathrm{t})
$$
$$
\therefore \quad \frac{d y}{d x}=\frac{a \sin t}{a(1+\cos t)}=\tan \frac{t}{2}
$$
$$
\begin{aligned}
\therefore \text { Length of subtangent } &=\frac{y}{d y / d x} \\
&=\frac{a(1-\cos t)}{\tan \frac{t}{2}} \\
&=2 a \sin \frac{t}{2} \cos \frac{t}{2} \\
&=a \sin t
\end{aligned}
$$
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