If x=sec θ-cos θ, y=sec ^n θ-cos ^n θ then ≤ft((d y)/(d x))^2 is equa…

$\frac{d y}{d \theta}=n \sec ^{n-1} \theta . \sec \theta . \tan \theta-n . \cos ^{n-1} \theta .(-\sin \theta)$

$=n {\left[\sec ^n \theta \frac{\sin \theta}{\cos \theta}+\cos ^{n-1} \theta . \sin \theta\right] }$

$\frac{n \sin \theta}{\cos \theta}\left[\sec ^n \theta+\cos ^n \theta\right]$

$n \tan \theta\left(\sec ^n \theta+\cos ^n \theta\right)$

$\frac{d x}{d \theta}=\sec \theta \tan \theta+\sin \theta=\sec \theta \frac{\sin \theta}{\cos \theta}+\sin \theta$

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

∴  $\frac{d y}{d x}=\frac{n \tan \theta\left(\sec ^n \theta+\cos ^n \theta\right)}{\tan \theta(\sec \theta+\cos \theta)}$

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

∴  $\left(\frac{d y}{d x}\right)^2=\frac{n^2\left(\sec ^n \theta+\cos ^n \theta\right)^2}{(\sec \theta+\cos \theta)^2}$

Hence (1) is correct answer.