Formula used:\(tan(A-B) = \frac{tan\ A - tan \ B}{1 + tanA\ tanB}\)\( tan\ 2? = \frac{2tan?}{1-tan^2?}\)Calculation:Given that, 2 tan ? = 1? tan ? = 1/2 ------(1)Also, we have \(? + ? = \frac{?}{4}\)? \(? = \frac{?}{4}-?\)\(? tan\ ? = tan(\frac{?}{4}-?)\)\(? tan\ ? = tan(1 -?)\) [? tan(?/4) = 1]By using the formula given above\(? tan\ ? = \frac{1 -tan?}{1+tan?}\)Put the value of tan? from equation (1)\(? tan\ ? = \frac{1 -\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3}\)? tan ? = 1/3 -----(2)We know that, \( tan\ 2? = \frac{2tan?}{1-tan^2?}??\)\( ? tan\ 2? = \frac{2(\frac{1}{3})}{1-(\frac{1}{3})^2}=\frac{\frac{2}{3}}{1-\frac{1}{9}}\)? tan 2? = \(\frac{3}{4}\)