Question

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

Answer 1

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

$=\frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1-\left(\frac{m}{m+1}\right)\left(\frac{1}{2m+1}\right)}$

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

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

$=1$

$\Rightarrow \tan (\alpha +\beta )=1$ since $\tan \frac{\pi }{4}=1$

$\Rightarrow \alpha +\beta =\frac{\pi }{4}$