Question

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

• A. $ 0 $
• B. $ 1 $
• C. $ -1 $
• D. $ 2 $

Answer 1

Answer: (C)

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