$\tan \alpha =2,\tan \beta =\frac{1}{3}$ با توجه به رابطهٔ $\tan (a\pm b)=\frac{\tan a\pm \tan b}{1\mp \tan a\tan b}$ داریم: $1)\tan (2\alpha -\beta )=\frac{\tan 2\alpha -\tan \beta }{1+\tan 2\alpha \tan \beta }$ $2)\tan 2\alpha =\frac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }\Rightarrow \tan 2\alpha =\frac{2\times 2}{1-{{2}^{2}}}=\frac{-4}{3}$ $\xrightarrow{(1)}\tan (2\alpha -\beta )=\frac{-\frac{4}{3}-\frac{1}{3}}{1+\left( -\frac{4}{3} \right)\left( \frac{1}{3} \right)}=\frac{-\frac{5}{3}}{\frac{5}{9}}=-3$