Question

If $\tan \alpha =\frac{m}{m+1}$ and $\tan \beta =\frac{1}{2m+1}$ then prove that $\alpha +\beta =\frac{\pi }{4}$

Answer 1

Given that: $\tan \alpha =\frac{m}{m+1}$ and $\tan \beta =\frac{1}{2m+1}$

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

$=\frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1-\frac{m}{m+1}\times \frac{1}{2m+1}}$

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

$=\frac{2m^2+2m+1}{2m^2+3m+1-m}$

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

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