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If $x= \sec \theta -\cos \theta $ and $y=\sec^n \theta -\cos ^n \theta$ then $\bigg(\large\frac{dy}{dx}\bigg)^2=$

$(a)\;\frac{n^2(y^2+2)}{x^2+4} \\ (b)\;\frac{(n^2+2)(y^2+4)}{(x^2+4)} \\ (c)\;\frac{x^2+4}{(y^2+4)} (n^2+2) \\ (d)\;\frac{y^2+4}{x^2+4}(n^2) $

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$y= \sec ^n \theta - \cos ^n \theta$
$\large\frac{dy}{d \theta}$$= n \sec^{n-1} \theta . \sec\theta -n \cos ^{n-1} \theta (-\sin \theta)$
$\large\frac{dy}{d\theta}$$ = n \sec^n \theta + n \cos^n \theta ^{-1} \sin \theta$
$\large\frac{dn}{d\theta}$$=\sec \theta +\sin \theta$
$\large\frac{dy}{dn} =\large\frac{n \tan \theta \bigg[\sec^n \theta + \cos ^n \theta\bigg]}{\sec \theta \tan \theta +\sin \theta}$
$\qquad= \large\frac{n \;\tan \theta [ \sec^n \theta +\cos ^n \theta ]}{\tan \theta [\sec \theta + \cos \theta ]}$
$\bigg(\large\frac{dy}{dx} \bigg)^2$$ =\large\frac{n^2( \sec^{2n} \theta+ \cos ^{2n} \theta +2 )}{ \sec ^2 \theta +\cos ^2 \theta +2}$
$x^2+4=\sec^2 \theta +\cos ^2 \theta +2$
$y^2 +4 =\sec^{2x} \theta +\cos ^{2x} \theta +2 $
So, $\bigg(\large\frac {dy}{dx}\bigg)^2 =\large\frac{x^2(y^2+4)}{x^2+4}$
or smart so,
Put $n=0$
$y=0$
so, $\large\frac{dy}{dx}$$=0$
Check option correspondingly B and C rejected
Now $n=1$
$y=x$
$\large\frac{dy}{dx}$$=1$
Hence d is the correct answer.
answered Jan 31, 2014 by meena.p
 

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