Solve this

as $x=$ cost

Differentiating it with respect to $t$,

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\cos \mathrm{t})$

$\frac{d x}{d t}=-\sin t$             .....(1)

And, $y=\sin t$

Differentiating it with respect to $t$,

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\sin t)$

$\frac{d y}{d t}=\cos t \ldots \ldots(2)$

Dividing equation (2) by (1),

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{\cos \mathrm{t}}{-\sin \mathrm{t}}$

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

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=-\cot \left(\frac{2 \pi}{3}\right)$

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=-\cot \left(\pi-\frac{\pi}{3}\right)$

$=-\left[-\cot \left(\frac{\pi}{3}\right)\right]$

$=\cot \left(\frac{\pi}{3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{3}}$