Question

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

• A. $\frac{n^2(y^2+4)}{x^2+4}$
• B. $\frac{n^2(y^2-4)}{x^2}$
• C. $n\frac{(y^2-4)}{x^2-4}$
• D. ${\left(\frac{ny}{x}\right)}^2-4$

Answer 1

The correct option is A $\frac{n^2(y^2+4)}{x^2+4}$

Given, $x=\sec \theta -\cos \theta$ and $y={\sec }^n\theta -{\cos }^n\theta$

Differentiate given equations with respect to $\theta$, we get

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

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

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

$\therefore$, $\frac{dy}{dx}=\frac{n\tan \theta ({\sec }^n\theta +{\cos }^n\theta )}{\tan \theta (\sec \theta +\cos \theta )}$

$=\frac{n({\sec }^n\theta +{\cos }^n\theta )}{(\sec \theta +\cos \theta )}$
Square of the given differential would be ${\left(\frac{dy}{dx}\right)}^2=\frac{n^2{({\sec }^n\theta +{\cos }^n\theta )}^2}{{(\sec \theta +\cos \theta )}^2}$

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

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