It is easy to see that $a_n > 0$. Let $a_n=\tan{\theta_n}$, then $\theta_n\in\Big(0, \frac{\pi}{2}\Big)$. $$a_n=\frac{\sqrt{1+\tan^2\theta_{n-1}}-1}{\tan\theta_{n-1}}=\frac{1-\cos\theta_{n-1}}{\sin\theta_{n-1}}=\tan\frac{\theta_{n-1}}{2}$$ It follows $$\tan\theta_{n}=\tan\frac{\theta_{n-1}}{2}\implies \theta_n = \frac{\theta_{n-1}}{2}$$ $$a_0=1 \implies \theta_{0}=\frac{\pi}{4}$$ Hence, $\{\theta_n\}$ is a geometric sequence whose initial term $\theta_0=\frac{\pi}{4}$ and common ratio is $\frac{1}{2}$. This implies $$\theta_n=\Big(\frac{1}{2}\Big)^n \cdot \frac{\pi}{4}=\frac{\pi}{2^{n+2}}$$ Given $\theta_n\in\Big(0, \frac{\pi}{2}\Big)$, we have $$a_n =\tan\theta_{n} > \theta_n= \frac{\pi}{2^{n+2}}$$