\[\boxed{\mathbf{618}\mathbf{.}}\]
\[\cos\left( \text{arctg\ a} \right) = \frac{1}{\sqrt{1 + a^{2}}}\]
\[\cos^{2}\left( \text{arctg\ a} \right) = \frac{1}{1 + a^{2}}\]
\[\frac{1}{1 + tg^{2}\left( \text{arctg\ a} \right)} = \frac{1}{1 + a^{2}}\]
\[\frac{1}{1 + a^{2}} = \frac{1}{1 + a^{2}}\]
\[Что\ и\ требовалось\ доказать.\]