\(tan 2\alpha=\frac{2tan\alpha}{1-tan^2\alpha}\) = \(\frac{2(\frac{1}{3})}{1-(\frac{1}{3})^2}=\frac{\frac{2}{3}}{\frac{8}{8}}=\frac{3}{4}\)
\(tan(2\alpha+\beta)= \frac{tan2\alpha+tan\beta}{1-tan2\alpha+tan\beta}\)
\(=\frac{\frac{3}{4}+\frac{1}{7}}{1-(\frac{3}{4})(\frac{1}{7})}\) \(=\frac{\frac{21+4}{28}}{\frac{28-3}{28}}=1\)
∴ 2α + β = 45° = π/4
