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

Given,
\(x =\sec \theta, y=\tan \theta \)
\( \frac{d x}{d \theta} =\sec \theta \tan \theta, \frac{d y}{d \theta}=\sec ^{2} \theta \)
\( \therefore \quad \frac{d y}{d x} =\frac{\sec ^{2} \theta}{\sec \theta \tan \theta}=\operatorname{cosec} \theta \)
\( \text { Now, } \frac{d^{2} y}{d x^{2}} =\frac{d}{d x}\left(\frac{d y}{d x}\right) \)
\( =\frac{d}{d \theta}(\operatorname{cosec} \theta) \frac{d \theta}{d x} \)
\( =-\operatorname{cosec} \theta \cot \theta \times \frac{1}{\sec 0 \tan 0} \)
\( =-\frac{1}{\tan ^{3} \theta} \)
\( \text { At } \theta=\frac{\pi}{4}, \left(\frac{d^{2} y}{d x^{2}}\right)_{\theta=\frac{\pi}{4}}=-\frac{1}{\left(\tan \frac{\pi}{4}\right)^{3}}=-1\)