If $\tan \alpha=\frac{m}{m+1}, \tan \beta=\frac{1}{2 m+1}$, then $\alpha+\beta$ equal to

We know that,

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

$\Rightarrow \tan (\alpha+\beta)=\frac{\frac{m}{m+1}+\frac{1}{2 m+1}}{1-\frac{m}{(m+1)} \cdot \frac{1}{(2 m+1)}}$

$\because \tan \alpha=\frac{m}{m+1}$

$\tan \beta=\frac{1}{2 m+1}=\frac{2 m^{2}+m+m+1}{\frac{(m+1)(2 m+1)}{2 m^{2}+3 m+1-m}}$

$=\frac{2 m^{2}+2 m+1}{2 m^{2}+2 m+1}=1$

$\Rightarrow \tan (\alpha+\beta)=1$

$\tan (\alpha+\beta)=\tan \frac{\pi}{4}$

$\therefore \alpha+\beta=\frac{\pi}{4}$.