Question

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

• A. 0
• B. 2
• C. 4
• D. 9

Answer 1

Answer: (D)

We have, $x=\sec \theta -\cos \theta$
$\Rightarrow \frac{dx}{d\theta }=\sec \theta \cdot \tan \theta +\sin \theta$
$\therefore \frac{dx}{d\theta }=\tan \theta \cdot (\sec \theta +\cos \theta )$
Also $y={\sec }^3\theta -{\cos }^3\theta$ )
$\Rightarrow \frac{dy}{d\theta }=3{\sec }^2\theta \cdot \sec \theta \cdot \tan \theta +3{\cos }^2\theta \cdot \sin \theta =3\tan \theta \left({\sec }^3\theta +{\cos }^3\theta \right)$
So, ${\left(\frac{dy}{dx}\right)}^2={\left(\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}\right)}^2=\frac{9{\tan }^2\theta {\left({\sec }^3\theta +{\cos }^3\theta \right)}^2}{{\tan }^2\theta (\sec \theta +\cos \theta {)}^2}$
So, ${\left(\frac{dy}{dx}\right)}_{x=0}^2\Rightarrow {\left(\frac{dy}{dx}\right)}_{\text{at }\theta =n\pi (n\in I)}^2=9$