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