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If $x=\sec \theta-\cos \theta, y=\sec ^{n} \theta-\cos ^{n} \theta$, then $\left(x^{2}+4\right)\left(\frac{d y}{d x}\right)^{2}$ is equal to
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The correct answer is:
$n^{2}\left(y^{2}+4\right)$
$x=\sec \theta-\cos \theta$
$\Rightarrow \frac{d x}{d \theta}=\sec \theta \tan \theta+\sin \theta$
$y=\sec ^{n} \theta-\cos ^{n} \theta$
$\Rightarrow \frac{d y}{d \theta}=n \sec ^{n-1} \theta \sec \theta \tan \theta+n \cos ^{n-1} \theta \sin \theta$
$\therefore \frac{d y}{d x}=n \frac{\left(\sec ^{n} \theta \tan \theta+\cos ^{n-1} \theta \sin \theta\right)}{(\sec \theta \tan \theta+\sin \theta)}$
$\Rightarrow \frac{d y}{d x}=n \frac{\left(\sec ^{n} \theta+\cos ^{n} \theta\right) \tan \theta}{(\sec \theta+\cos \theta) \tan \theta}$
$$
\begin{array}{l}
\Rightarrow \frac{d y}{d x}=\frac{n\left(\sec ^{n} \theta+\cos ^{n} \theta\right)}{(\sec \theta+\cos \theta)} \\
\Rightarrow\left(\frac{d y}{d x}\right)^{2}=\frac{n^{2}\left\{\left(\sec ^{n} \theta-\cos ^{n} \theta\right)^{2}+4\right\}}{(\sec \theta-\cos \theta)^{2}+4} \\
\Rightarrow\left(\frac{d y}{d x}\right)^{2}=\frac{n^{2}\left(y^{2}+4\right)}{x^{2}+4} \\
\Rightarrow\left(x^{2}+4\right)\left(\frac{d y}{d x}\right)^{2}=n^{2}\left(y^{2}+4\right)
\end{array}
$$
$\Rightarrow \frac{d x}{d \theta}=\sec \theta \tan \theta+\sin \theta$
$y=\sec ^{n} \theta-\cos ^{n} \theta$
$\Rightarrow \frac{d y}{d \theta}=n \sec ^{n-1} \theta \sec \theta \tan \theta+n \cos ^{n-1} \theta \sin \theta$
$\therefore \frac{d y}{d x}=n \frac{\left(\sec ^{n} \theta \tan \theta+\cos ^{n-1} \theta \sin \theta\right)}{(\sec \theta \tan \theta+\sin \theta)}$
$\Rightarrow \frac{d y}{d x}=n \frac{\left(\sec ^{n} \theta+\cos ^{n} \theta\right) \tan \theta}{(\sec \theta+\cos \theta) \tan \theta}$
$$
\begin{array}{l}
\Rightarrow \frac{d y}{d x}=\frac{n\left(\sec ^{n} \theta+\cos ^{n} \theta\right)}{(\sec \theta+\cos \theta)} \\
\Rightarrow\left(\frac{d y}{d x}\right)^{2}=\frac{n^{2}\left\{\left(\sec ^{n} \theta-\cos ^{n} \theta\right)^{2}+4\right\}}{(\sec \theta-\cos \theta)^{2}+4} \\
\Rightarrow\left(\frac{d y}{d x}\right)^{2}=\frac{n^{2}\left(y^{2}+4\right)}{x^{2}+4} \\
\Rightarrow\left(x^{2}+4\right)\left(\frac{d y}{d x}\right)^{2}=n^{2}\left(y^{2}+4\right)
\end{array}
$$
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