If \( \tan A=\frac{1}{2} \) and \( \tan B=\frac{1}{3} \), Find \( \tan(2A+B) \)
Question
If
\[ \tan A=\frac{1}{2},\qquad \tan B=\frac{1}{3}, \]
then
\[ \tan(2A+B) \]
is equal to
(a) 1
(b) 2
(c) 3
(d) 4
Solution
First find \(\tan 2A\):
\[ \tan 2A = \frac{2\tan A}{1-\tan^2 A} = \frac{2\left(\frac12\right)} {1-\left(\frac12\right)^2} \]
\[ = \frac{1}{1-\frac14} = \frac{1}{\frac34} = \frac43 \]
Now use
\[ \tan(2A+B) = \frac{\tan2A+\tan B} {1-\tan2A\tan B} \]
\[ = \frac{\frac43+\frac13} {1-\frac43\cdot\frac13} \]
\[ = \frac{\frac53} {1-\frac49} = \frac{\frac53} {\frac59} \]
\[ = 3 \]
Final Answer
\[ \boxed{3} \]
Hence, the correct option is (c) 3.