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

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

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

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

$=\tan \theta (\sec \theta +\cos \theta )$

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

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

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