$\begin{align}  & (1-\sqrt{3}\tan {{10}^{\circ }})(\frac{\sqrt{3}}{3}+\tan {{70}^{\circ }}) \\  & =\frac{\sqrt{3}}{3}+\tan {{70}^{\circ }}-\tan {{10}^{\circ }}-\sqrt{3}\tan {{10}^{\circ }}\tan {{70}^{\circ }}\,\,(1) \\ \end{align}$ از طرفی داريم: $\begin{align}  & \tan {{60}^{\circ }}=\tan ({{70}^{\circ }}-{{10}^{\circ }})=\frac{\tan {{70}^{\circ }}-\tan {{10}^{\circ }}}{1+\tan {{70}^{\circ }}\tan {{10}^{\circ }}} \\  & \Rightarrow \sqrt{3}=\frac{\tan {{70}^{\circ }}-\tan {{10}^{\circ }}}{1+\tan {{70}^{\circ }}\tan {{10}^{\circ }}} \\  & \Rightarrow \sqrt{3}+\sqrt{3}\tan {{70}^{\circ }}\tan {{10}^{\circ }}=\tan {{70}^{\circ }}-\tan {{10}^{\circ }} \\  & \Rightarrow \tan {{70}^{\circ }}-\tan {{10}^{\circ }}-\sqrt{3}\tan {{70}^{\circ }}\tan {{10}^{\circ }}=\sqrt{3}\,\,\,(2) \\  & \xrightarrow{(1),(2)}\frac{\sqrt{3}}{3}+\sqrt{3}=\frac{4\sqrt{3}}{3} \\ \end{align}$