If $x=a\left\{\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right\}$ and $y=a \sin \theta$, then…
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
- $\cot \theta$
- $\tan \theta$
- $\sin \theta$
- $\cos \theta$
Solution
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}
$
Asked in: AP EAMCET 2008
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