If tan A = 1/2 and tan B = 1/3, then what is the value of tan (2A + B)?

Given : $$tanA=\frac{1}{2}$$ and $$tanB=\frac{1}{3}$$

Now, $$tan2A=\frac{2tanA}{1-tan^2A}$$

=> $$tan2A=(2\times\frac{1}{2})\div(1-\frac{1}{4})$$

=> $$tan2A=1\div\frac{3}{4}$$

=> $$tan2A=\frac{4}{3}$$

To find : $$tan(2A+B)$$

= $$\frac{tan2A+tanB}{1-tan2A.tanB}$$

= $$(\frac{4}{3}+\frac{1}{3})\div(1-(\frac{4}{3})(\frac{1}{3}))$$

= $$(\frac{5}{3})\div(1-\frac{4}{9})$$

= $$(\frac{5}{3})\div(\frac{9-4}{9})$$

= $$\frac{5}{3}\times\frac{9}{5}=3$$

=> Ans - (B)