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)

Given, $x=a{\cos }^3\theta$ and $y=a{\sin }^3\theta$
On differentiating both sides w.r.t. $\theta$, we get
$\frac{dx}{d\theta }=3a{\cos }^2\theta (-\sin \theta )$
and $\frac{dy}{d\theta }=3a{\sin }^2\theta (\cos \theta )$
Now, $\frac{dy/d\theta }{dx/d\theta }=\frac{3a{\sin }^2\theta (\cos \theta )}{3a{\cos }^2\theta (-\sin \theta )}$
$\Rightarrow \frac{dY}{dx}=-\tan \theta$
$\therefore 1+{\left(\frac{dy}{dx}\right)}^2=1+(-\tan \theta {)}^2$
$=1+{\tan }^2\theta$
$={\sec }^2\theta$