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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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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}$

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

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$x=\cos t (3-2\cot^2t)$
$ = 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$
answered Apr 14, 2014 by thanvigandhi_1
 

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