Question

If $x=\sec \theta-\cos \theta$ and $y=\sec ^3 \theta-\cos ^3 \theta$, then the value of $\left(\frac{d y}{d x}\right)^2$ 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, $\left(\frac{d y}{d x}\right)^2=\left(\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\right)^2=\frac{9 \tan ^2 \theta\left(\sec ^3 \theta+\cos ^3 \theta\right)^2}{\tan ^2 \theta(\sec \theta+\cos \theta)^2}$
So, $\left(\frac{d y}{d x}\right)_{x=0}^2 \Rightarrow\left(\frac{d y}{d x}\right)_{\text {at } \theta=n \pi(n \in I)}^2=9$