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AP EAPCET 20-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 65 of 160

Marks: +1, -0

If

x=\sec\theta-\cos\theta

y=\sec^{n}\theta-\cos^{n}\theta

(x^{2}+4)\left(\frac{dy}{dx}\right)^{2}=

_______

$n(y+4)$
$n^{2}(y^{2}+4)$
$n(y+2)$
$n^{2}(y^{2}+2)$

Solution:

x=\sec\theta-\cos\theta

, then

\frac{dx}{d\theta}=\sec\theta\tan\theta+\sin\theta=\tan\theta(\sec\theta+\cos\theta)

y=\sec^{n}\theta-\cos^{n}\theta

, then

\frac{dy}{d\theta}=n\sec^{n-1}\theta\sec\theta\tan\theta+n\cos^{n-1}\theta\sin\theta

=n\tan\theta(\sec^{n}\theta+\cos^{n}\theta)

\therefore\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

=\frac{n\tan\theta(\sec^{n}\theta+\cos^{n}\theta)}{\tan\theta(\sec\theta+\cos\theta)}

=\frac{n(\sec^{n}\theta+\cos^{n}\theta)}{\sec\theta+\cos\theta}

\left(\frac{dy}{dx}\right)^{2}=\frac{n^{2}(\sec^{n}\theta+\cos^{n}\theta)^{2}}{(\sec\theta+\cos\theta)^{2}}

=\frac{n^{2}\{(\sec^{n}\theta-\cos^{n}\theta)^{2}+4\}}{(\sec\theta-\cos\theta)^{2}+4}

=\frac{n^{2}(y^{2}+4)}{x^{2}+4}

\Rightarrow(x^{2}+4)\left(\frac{dy}{dx}\right)^{2}=n^{2}(y^{2}+4)

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