Question

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

• A. $ta{n}^2\theta$
• B. $co{t}^2\theta$
• C. $se{c}^2\theta$
• D. $cose{c}^2\theta$
• E. $se{c}^3\theta$

Answer 1

Answer: (C)

$x={\cos }^3\theta ,y={\sin }^3\theta$ On differentiating w.r.t. $\theta$ respectively $\frac{dx}{d\theta }=-3{\cos }^2\theta \sin \theta$ and $\frac{dy}{d\theta }=3{\sin }^2\theta \cos \theta$ Now, $\frac{dy}{dx}=-\frac{3{\sin }^2\theta \cos \theta }{3{\cos }^2\theta \sin \theta }=-\tan \theta$ $\therefore$ $1+{\left(\frac{dy}{dx}\right)}^2=1+{\tan }^2\theta ={\sec }^2\theta$