#### Question

If \[x = \cos t \text{ and y } = \sin t,\] prove that \[\frac{dy}{dx} = \frac{1}{\sqrt{3}} \text { at } t = \frac{2 \pi}{3}\] ?

#### Solution

\[\text{ We have, x } = \cos t \text{ and y } = \sin t\]

\[\Rightarrow \frac{dx}{dt} = \frac{d}{dt}\left( \cos t \right) \text { and } \frac{dy}{dt} = \frac{d}{dt}\left( \sin t \right)\]

\[ \Rightarrow \frac{dx}{dt} = - \sin t \text{ and } \frac{dy}{dt} = \cos t \]

\[ \therefore \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\cos t}{- \sin t} = - \cot t \]

\[\text{ Now,} \left( \frac{dy}{dx} \right)_{t = \frac{2\pi}{3}} = - \cot \left( \frac{2\pi}{3} \right) = \frac{1}{\sqrt{3}} \]

\[ \Rightarrow \frac{dx}{dt} = - \sin t \text{ and } \frac{dy}{dt} = \cos t \]

\[ \therefore \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\cos t}{- \sin t} = - \cot t \]

\[\text{ Now,} \left( \frac{dy}{dx} \right)_{t = \frac{2\pi}{3}} = - \cot \left( \frac{2\pi}{3} \right) = \frac{1}{\sqrt{3}} \]

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Solution If X = Cos T and Y = Sin T , Prove that D Y D X = 1 √ 3 at T = 2 π 3 ? Concept: Simple Problems on Applications of Derivatives.