Question

If $x=sec\theta$ , $y=tan\theta$ , then the value of $\frac{d^{2}y}{d{x}^2}$ at $\theta =\frac{\pi }{4}$ is

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (C)

Given, $x=\sec \theta ,y=\tan \theta$
$\frac{dx}{d\theta }=\sec \theta \tan \theta ,\frac{dy}{d\theta }={\sec }^2\theta$
$\therefore \frac{dy}{dx}=\frac{{\sec }^2\theta }{\sec \theta \tan \theta }=cosec\theta$
Now, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$
$=\frac{d}{d\theta }(cosec\theta )\frac{d\theta }{dx}$
$=-cosec\theta \cot \theta \times \frac{1}{\sec \theta \tan \theta }$
$=-\frac{1}{{\tan }^3\theta }$
At $\theta =\frac{\pi }{4},{\left(\frac{d^{2}y}{d{x}^2}\right)}_{\theta =\frac{\pi }{4}}=-\frac{1}{{\left(\tan \frac{\pi }{4}\right)}^3}=-1$