# If $x = \cos\: t ( 3-2 \: \cos^2 t)\: and \: y = \sin \: t ( 3-2 \sin^2\: t)$ find the value of $\large\frac{dy}{dx}$ at $t = \large\frac{\pi}{4}$

$(a)\; sint\qquad(b)\; 3cost-sint\qquad(c)\; sin2t .sint\qquad(d)\; 3cost-3sin2t .sint$

$x=\cos t (3-2\cot^2t)$
$= 3 \cos t - 2 \cos^3t$
$\large\frac{dx}{dt}$$= - 3 \sin t-6 \cos^2t(-\sin t) = 6 \sin t \cos^2t-3 \sin t = 3 \times 2 \sin t \cos t. \cos t-3 \sin t = 3\sin 2t \cos t - 3 \sin t y= \sin t (3-2 \sin^2t) y=3 \sin t-2 \sin^3 t \large\frac{dy}{dx}$$ = 3 \cos t-6 \sin^2t ( \cos t)$
$3 \cos t - 3 \times 2 \sin t \cos t.\sin t$
$= 3 \cos t - 3 \sin 2t. \sin t$