Question

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

• A. ${\tan }^2\theta$
• B. ${\cot }^2\theta$
• C. ${\sec }^2\theta$
• D. ${\text{cosec}}^2\theta$

Answer 1

The correct option is C ${\sec }^2\theta$

$\Rightarrow \frac{dx}{d\theta }=3{\cos }^2\theta (-\sin \theta )=-3{\cos }^2\theta \sin \theta$ ------(1)

and

$\Rightarrow \frac{dy}{d\theta }=3{\sin }^2\theta \cos \theta$ -----(2)

Hence, $1+{\left(\frac{dy}{dx}\right)}^2=1+{\left(\frac{dy/d\theta }{dx/d\theta }\right)}^2=1+\frac{{\sin }^2\theta }{{\cos }^2\theta }={\sec }^2\theta$