Question

If $tanA=\frac{1}{2}$, $tanB=\frac{1}{3}$, then $tan\left(2A+B\right)$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

Given, $tanA=\frac{1}{2}, tanB=\frac{1}{3}\quad\ldots\left(i\right)$
Now, $tan\left(2A+B\right)=\frac{tan\,2A+tan\,B}{1-tan\,2A\,tan\,B}$
$=\frac{\frac{2\,tan\,A}{1-tan^{2}\,A}+tan\,B}{1-\frac{2\,tan\,A}{1-tan^{2}\,A}\times tan\,B}$
$=\frac{\frac{4}{3}+\frac{1}{3}}{1-\frac{4}{3}\times\frac{1}{3}}=3$