Let $x_n = \tan\alpha_n$ and $y_n=\tan\beta_n$ for all $n \ge 1$ where $\alpha_1=\beta_1=60^\circ$ and $0 < \alpha_n, \beta_n < 90^\circ$ because all $x_n$ and $y_n$ are obviously positive. Then, \begin{align*} \tan\alpha_{n+1} &= \tan\alpha_n +\sqrt{1+\tan^2\alpha_n}\\ &= \tan\alpha_n + \sec\alpha_n\\ &=\frac{\sin\alpha_n + 1}{\cos\alpha_n}\\ &=\frac{1-\cos(90^\circ +\alpha_n)}{\sin(90^\circ+\alpha_n)}\\ &=\tan\Big(\frac{90^\circ+\alpha_n}{2}\Big)\\ &=\tan\Big(90^\circ-\frac{90^\circ-\alpha_n}{2}\Big) \end{align*} The last step holds because of \myJustRefP{eq_tan_half}. $$\therefore\quad \alpha_n=90^\circ - \frac{90^\circ -\alpha_{n-1}}{2}=\cdots=90^\circ -\frac{90^\circ-\alpha_1}{2^{n-1}}=90^\circ -\frac{30^\circ}{2^{n-1}}\quad (n > 1)$$ Let $\theta_n=\frac{30^\circ}{2^{n-1}}$. Then, $$x_{n}=\tan\alpha_{n}=\tan(90^\circ-\theta_n)=\cot\theta_n\qquad(n>1)$$ Similarly, it can be shown that $$y_n=\tan 2\theta_n\qquad(n>1)$$ This implies that for $n > 1$: $$x_ny_n=\cot\theta_n\cdot\tan 2\theta_n=\frac{1}{\tan\theta_n}\cdot\frac{2\tan\theta_n}{1-\tan^2\theta_n}=\frac{2}{1-\tan^2\theta_n}$$ Because $0 < \theta_n < 30^\circ$, therefore $0 < \tan^2\theta_n < \frac{1}{3}$. Substituting this range to the relation above immediately leads to the conclusion that $$2 < x_ny_n < 3$$