If $x=\cos t\left(3-2 \cos ^{2} t\right), y=\sin t\left(3-2 \sin ^{2} t\right)$ find the value of $\frac{d y}{d x}$ at $t=\frac{\pi}{4}$
$x=\cos t\left(3-2 \cos ^{2} t\right)$ and $y=\sin t\left(3-2 \sin ^{2} t\right)$
$\Rightarrow \frac{d x}{d t}=-\sin t\left(3-2 \cos ^{2} t\right)+\cos t(4 \cos t \sin t)$ and $\frac{d y}{d t}=\cos t\left(3-2 \sin ^{2} t\right)+\sin t(-4 \sin t \cos t)$
$\Rightarrow \frac{d x}{d t}=-3 \sin t+6 \sin t \cos ^{2} t$ and $\frac{d y}{d t}=3 \cos t-6 \sin ^{2} t \cos t$
$\Rightarrow \frac{d x}{d t}=-3 \sin t\left(1-2 \cos ^{2} t\right)$ and $\frac{d y}{d t}=3 \cos t\left(1-2 \sin ^{2} t\right)$
$\Rightarrow \frac{d x}{d t}=3 \sin t \cos 2 t$ and $\frac{d y}{d t}=3 \cos t \cos 2 t$
$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{3 \cos t \cos 2 t}{3 \sin t \cos 2 t}=\cot t$
Now, $\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{4}}=\cot \frac{\pi}{4}=1$