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If the parametric equation of curve is given by $x=\cos \theta+\log \tan \frac{\theta}{2}$ and $y=\sin \theta$, then the points for which $\frac{d y}{d x}=0$ are given by
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Verified Answer
The correct answer is:
$\theta=n \pi, n \in z$
$x=\cos \theta+\log \tan \frac{\theta}{2}$
Oñ đifferrēñtiāting ̄ w.r.t. $\theta$, wē gẹt
$\frac{d x}{d \theta}=-\sin \theta+\frac{1}{\tan \frac{\theta}{2}} \times \sec ^{2} \frac{\theta}{2} \times \frac{1}{2}$
$\begin{aligned}
&=-\sin \theta+\frac{1}{\sin \theta} \\
&=\frac{1-\sin ^{2} \theta}{\sin \theta} \\
&=\frac{\cos ^{2} \theta}{\sin \theta} ...(i)
\end{aligned}$
Now, $y=\sin \theta$
On differentiating w.r.t. $\theta$,
$\frac{d y}{d \theta}=\cos \theta...(ii)$
$\begin{aligned}
\therefore \quad \frac{d y}{d x} &=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \\
&=\frac{\cos \theta}{\frac{\cos ^{2} \theta}{\sin \theta}}=\tan \theta
\end{aligned}$
[from Eqs. (i) and (ii)]
If $\frac{d y}{d x}=0$, then $\tan \theta=0$.
Hence, $\theta=n \pi, n \in Z$.
Oñ đifferrēñtiāting ̄ w.r.t. $\theta$, wē gẹt
$\frac{d x}{d \theta}=-\sin \theta+\frac{1}{\tan \frac{\theta}{2}} \times \sec ^{2} \frac{\theta}{2} \times \frac{1}{2}$
$\begin{aligned}
&=-\sin \theta+\frac{1}{\sin \theta} \\
&=\frac{1-\sin ^{2} \theta}{\sin \theta} \\
&=\frac{\cos ^{2} \theta}{\sin \theta} ...(i)
\end{aligned}$
Now, $y=\sin \theta$
On differentiating w.r.t. $\theta$,
$\frac{d y}{d \theta}=\cos \theta...(ii)$
$\begin{aligned}
\therefore \quad \frac{d y}{d x} &=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \\
&=\frac{\cos \theta}{\frac{\cos ^{2} \theta}{\sin \theta}}=\tan \theta
\end{aligned}$
[from Eqs. (i) and (ii)]
If $\frac{d y}{d x}=0$, then $\tan \theta=0$.
Hence, $\theta=n \pi, n \in Z$.
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