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If $x=\cos t$ and $y=\sin t$, prove that $\frac{d y}{d x}=\frac{1}{\sqrt{3}}$ at $t=\frac{2 \pi}{3}$


We have, $x=\cos t$ and $y=\sin t$

$\Rightarrow \frac{d x}{d t}=\frac{d}{d t}(\cos t)$ and $\frac{d y}{d t}=\frac{d}{d t}(\sin t)$

$\Rightarrow \frac{d x}{d t}=-\sin t$ and $\frac{d y}{d t}=\cos t$

$\therefore \frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\cos t}{-\sin t}=-\cot t$

Now, $\left(\frac{d y}{d x}\right)_{t=\frac{2 \pi}{3}}=-\cot \left(\frac{2 \pi}{3}\right)=\frac{1}{\sqrt{3}}$

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