Mathematics - Application of Derivative Question with Solution | TestHub

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