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If $x=a\left\{\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right\}$ and $y=a \sin \theta$, then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
$\tan \theta$
Given that,
$$
x=a\left(\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right) \text { and } y=a \sin \theta
$$
On differentiating w.r.t. $\theta$ respectively, we get
$$
\begin{aligned}
\frac{d x}{d \theta} & =a\left(-\sin \theta+\frac{1}{\tan \left(\frac{\theta}{2}\right)} \cdot \sec ^2 \frac{\theta}{2} \cdot \frac{1}{2}\right) \\
& =a\left(-\sin \theta+\frac{1}{\sin \theta}\right)=\frac{a \cos ^2 \theta}{\sin \theta}
\end{aligned}
$$
and $\frac{d y}{d \theta}=a \cos \theta$
$$
\begin{aligned}
\therefore \quad \frac{d y}{d x} & =\frac{d y / d \theta}{d x / d \theta}=\frac{a \cos \theta}{a \cos ^2 \theta / \sin \theta} \\
& =\tan \theta
\end{aligned}
$$
$$
x=a\left(\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right) \text { and } y=a \sin \theta
$$
On differentiating w.r.t. $\theta$ respectively, we get
$$
\begin{aligned}
\frac{d x}{d \theta} & =a\left(-\sin \theta+\frac{1}{\tan \left(\frac{\theta}{2}\right)} \cdot \sec ^2 \frac{\theta}{2} \cdot \frac{1}{2}\right) \\
& =a\left(-\sin \theta+\frac{1}{\sin \theta}\right)=\frac{a \cos ^2 \theta}{\sin \theta}
\end{aligned}
$$
and $\frac{d y}{d \theta}=a \cos \theta$
$$
\begin{aligned}
\therefore \quad \frac{d y}{d x} & =\frac{d y / d \theta}{d x / d \theta}=\frac{a \cos \theta}{a \cos ^2 \theta / \sin \theta} \\
& =\tan \theta
\end{aligned}
$$
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