Question:
Find $\frac{d y}{d x}$, when
If $x=\cos t\left(3-2 \cos ^{2} t\right)$ and $y=\sin t\left(3-2 \sin ^{2} t\right)$ find the value of $\frac{d y}{d x}$ at $t=\frac{\pi}{4}$
Solution:
considering the given functions,
$x=\cos t\left(3-2 \cos ^{2} t\right)$
$x=3 \cos t-2 \cos ^{3} t$
$\frac{d x}{d t}=-3 \sin t+6 \cos ^{2} t \sin t$ ......(1)
$\frac{d y}{d t}=3 \cos t+6 \sin ^{2} t \cos t$ .......(2)
$\frac{d y}{d x}=\frac{\frac{d y}{d x}}{\frac{d x}{d t}}=\frac{3 \cos t+6 \sin ^{2} t \cos t}{-3 \sin t+6 \cos ^{2} t \sin t} \mid$ from equation 1 and 2
$=\frac{\cot t\left(1-2\left(1-\cos ^{2} t\right)\right)}{\left(2 \cos ^{2} t-1\right)}=\cot t$
When $t=\frac{\pi}{4}$
$\frac{d y}{d x}=\cot \frac{\pi}{4}=1$