Answer:
[tex]x=a \cos t \implies \dfrac{dx}{dt}=-a \sin t[/tex]
[tex]y= a \sin t \implies \dfrac{dy}{dt}=a \cos t[/tex]
Using the chain rule:
[tex]\begin{aligned}\implies \dfrac{dy}{dx} & =\dfrac{dy}{dt} \times \dfrac{dt}{dx}\\\\& = a \cos t \times \dfrac{1}{-a \sin t}\\\\& = \dfrac{a \cos t}{-a \sin t}\\\\& = - \dfrac{\cos t}{\sin t}\\\\& = - \cot t\end{aligned}[/tex]
