Question

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

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

Answer 1

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

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