Let $\{x_i\}^{\infty}_{i=1}$ and $\{y_i\}^{\infty}_{i=1}$ be sequences of real numbers such that $x_1=y_1=\sqrt{3}$, $$x_{n+1}=x_n+\sqrt{1+x_n^2}\quad\text{and}\quad y_{n+1}=\frac{y_n}{1+\sqrt{1+y_n^2}}$$for all $n\geq 1$. Prove that $2<x_ny_n<3$ for all $n>1$.
Note that $x_1=\sqrt{3}=\cot\frac{\pi}{6}$ and $y_1=\sqrt{3}=\tan\frac{\pi}{3}$.
If $x_i=\cot \theta$ then $x_{i+1}=\cot\theta+\csc\theta=\frac{\cos\theta+1}{\sin\theta}=\frac{2\cos^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\cot\frac{\theta}{2}$.
If $y_i=\tan\theta$ then $y_{i+1}=\frac{\tan\theta}{1+\sec\theta}=\frac{\frac{\sin\theta}{\cos\theta}}{\frac{\cos\theta+1}{\cos\theta}}=\frac{\sin\theta}{\cos\theta+1}=\tan\frac{\theta}{2}$.
By induction, $x_n=\cot\frac{\pi}{2^{n-1}\cdot 6}$ and $y_n=\tan\frac{\pi}{2^{n-1}\cdot 3}$. Thus $x_ny_n=\cot\frac{\pi}{2^{n-1}\cdot 6}\cdot\tan\frac{\pi}{2^{n-1}\cdot 3}$.
Let $A=\frac{\pi}{2^{n-1}\cdot 6}$. We have $\tan 2A\cot A=\frac{\tan 2A}{\tan A}=\frac{\frac{2\tan A}{1-\tan^2 A}}{\tan A}=\frac{2}{1-\tan^2 A}>2$.
Since $\frac{1}{\sqrt{3}}=\tan\frac{\pi}{6}>\tan\frac{\pi}{2^{n-1}\cdot 6}=\tan A\implies \frac{1}{3}>\tan^2 A \implies 3>\frac{2}{1-\tan^2 A}$.
Thus $2<x_ny_n<3$ for all $n>1$.