Question

If $x=\sec \theta -\cos \theta ,y={\sec }^n\theta -{\cos }^n\theta$, then $(x^2+4){\left(\frac{dy}{dx}\right)}^2$ is equal to

• A. $n^2(y^2-4)$
• B. $n^2(4-y^2)$
• C. $n^2(y^2+4)$
• D. $Noneofthese$

Answer 1

Answer: (C)

$x=\sec \theta -\cos \theta$
$\Rightarrow \frac{dx}{d\theta }=\sec \theta \tan \theta +\sin \theta ,$
$y={\sec }^n\theta -{\cos }^n\theta$
$\Rightarrow \frac{dy}{d\theta }=n{\sec }^{n-1}\theta \sec \theta \tan \theta +n{\cos }^{n-1}\theta \sin \theta$
$\therefore \frac{dy}{dx}=n\frac{\left({\sec }^n\theta \tan \theta +{\cos }^{n-1}\theta \sin \theta \right)}{(\sec \theta \tan \theta +\sin \theta )}$
$\Rightarrow \frac{dy}{dx}=n\frac{\left({\sec }^n\theta +{\cos }^n\theta \right)\tan \theta }{(\sec \theta +\cos \theta )\tan \theta }$
$\Rightarrow \frac{dy}{dx}=\frac{n\left({\sec }^n\theta +{\cos }^n\theta \right)}{(\sec \theta +\cos \theta )}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}^2=\frac{n^2\left\{{\left({\sec }^n\theta -{\cos }^n\theta \right)}^2+4\right\}}{(\sec \theta -\cos \theta {)}^2+4}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}^2=\frac{n^2\left(y^2+4\right)}{x^2+4}$
$\Rightarrow \left(x^2+4\right){\left(\frac{dy}{dx}\right)}^2=n^2\left(y^2+4\right)$