Question

If $tan\alpha =\frac{m}{m+1}$ and $tan\beta =\frac{1}{2m+1}$, then a possible value of $(\alpha +\beta )$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{\pi }{4}$

$tan(\alpha +\beta )=\frac{tan\left(\alpha \right)+tan\left(\beta \right)}{1-tan\left(\alpha \right)tan\left(\beta \right)}$

$=\frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1-\left(\frac{m}{m+1}\right)\left(\frac{1}{2m+1}\right)}$
$=1$
$\therefore (\alpha +\beta )=\frac{\pi }{4}$ is a possible value.