Question

If $\tan A=\frac{1}{2},\tan B=\frac{1}{3},$ then tan (2A + B) is equal
(a) 1
(b) 2
(c) 3
(d) 4

Answer 1

$$\begin{gathered} \tan A=\frac{1}{2},\tan B=\frac{1}{3} \\ \mathrm{then}\tan \left(2A+B\right)=\frac{\tan 2A+\tan B}{1-\tan 2A\tan B} \\ \mathrm{Using}\mathrm{Identities}: \\ \left(\tan \left(a+b\right)=\frac{\tan a+\tan b}{1-\tan a\tan b}\right) \\ =\frac{\left({\displaystyle \frac{2\mathrm{ta}nA}{1-{\tan }^{2}A}}\right)+\tan B}{1-\left({\displaystyle \frac{2\tan A}{1-{\tan }^{2}A}}\right)\tan B}\left(\mathrm{using}\mathrm{identity}\tan 2a=\frac{2\tan a}{1-{\tan }^{2}a}\right) \\ =\frac{{\displaystyle \frac{2\left({\displaystyle \frac{1}{2}}\right)}{1-{\left({\displaystyle \frac{1}{2}}\right)}^2}+\frac{1}{3}}}{1-\frac{2\left({\displaystyle \frac{1}{2}}\right)}{1-{\left({\displaystyle \frac{1}{2}}\right)}^2}\times \frac{1}{3}} \\ =\frac{{\displaystyle \frac{4}{3}}+{\displaystyle \frac{1}{3}}}{1-{\displaystyle \frac{4}{3}}\times {\displaystyle \frac{1}{3}}} \\ =\frac{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{1-{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$9$}\right.}} \\ =\frac{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{{\displaystyle \frac{9-4}{9}}}=\frac{5}{3}\times \frac{9}{5} \\ \therefore \tan \left(2A+B\right)=3 \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{option}\mathrm{C}. \end{gathered}$$