Question

If $x=\sec \theta -\cos \theta$ and $y={\sec }^3\theta -{\sec }^3\theta -{\cos }^3\theta ,$ then the value of ${\left(\frac{dy}{dx}\right)}^2$ at $x=0,$ is

• A. $0$
• B. $2$
• C. $4$
• D. $1$

Answer 1

The correct option is A $0$

$x=\sec \theta -\cos \theta y={\sec }^3\theta -{\cos }^3\theta \frac{dx}{d\theta }=\sec \theta \tan \theta +\sin \theta \frac{dy}{d\theta }=3{\sec }^2\theta .\sec \theta \tan \theta +3{\cos }^2\theta .\sin \theta =3{\sec }^3\theta \tan \theta +3{\cos }^2\theta \sin \theta \frac{dy}{dx}=\frac{dy}{d\theta }/\frac{dx}{d\theta }=\frac{3{\cos }^2\theta \sin \theta +3{\sec }^3\theta \tan \theta }{\sec \theta \tan \theta +\sin \theta }\frac{dy}{dx}{]}_{x=0}=\frac{3\times 1\times 0+3\times 0\times 1}{1\times 0+0}=0$