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Question
\[\text { If }x = \cos t\left( 3 - 2 \cos^2 t \right), y = \sin t\left( 3 - 2 \sin^2 t \right) \text { find the value of } \frac{dy}{dx}\text{ at }t = \frac{\pi}{4}\] ?
Solution
\[x = \cos t\left( 3 - 2 \cos^2 t \right) \text { and y } = \sin t\left( 3 - 2 \sin^2 t \right)\]
\[ \Rightarrow \frac{dx}{dt} = - \sin t\left( 3 - 2 \cos^2 t \right) + \cos t\left( 4\cos t\sin t \right) \text { and } \frac{dy}{dt} = \cos t\left( 3 - 2 \sin^2 t \right) + \sin t\left( - 4\sin t\cos t \right)\]
\[ \Rightarrow \frac{dx}{dt} = - 3\sin t + 6\sin t \cos^2 t \text { and } \frac{dy}{dt} = 3\cos t - 6 \sin^2 t\cos t\]
\[ \Rightarrow \frac{dx}{dt} = - 3\sin t\left( 1 - 2 \cos^2 t \right) \text { and } \frac{dy}{dt} = 3\cos t\left( 1 - 2 \sin^2 t \right)\]
\[ \Rightarrow \frac{dx}{dt} = 3\sin t\cos2t \text { and } \frac{dy}{dt} = 3\cos t\cos2t\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3\cos t\cos2t}{3\sin t\cos2t} = \cot t\]
\[\text { Now,} \left( \frac{dy}{dx} \right)_{t = \frac{\pi}{4}} = \cot\frac{\pi}{4} = 1\]
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