Question

If $x=a{\cos }^3\theta$ and $y=a{\sin }^3\theta$, then $1+{\left(\frac{dy}{dx}\right)}^2$ is

• A. $\tan \theta$
• B. ${\tan }^2\theta$
• C. $1$
• D. ${\sec }^2\theta$
• E. $\sec \theta$

Answer 1

Answer: (D)

${\sec }^2\theta$

Given, $x=a{\cos }^3\theta ,y=a{\sin }^3\theta$

$\frac{dx}{d\theta }=3a{\cos }^2\theta .(-\sin \theta )$

and $\frac{dy}{d\theta }=3a{\sin }^2\theta .\cos \theta$

Therefore, $\frac{dy}{dx}=\frac{3a{\sin }^2\theta .\cos \theta }{-3a{\cos }^2\theta .\sin \theta }$ $=-\tan \theta$