Question

If $tan\alpha$ $=$ $\frac{m}{m+1}$ and $tan\beta$ $=$ $\frac{1}{1+2m}$ then $tan(\alpha +\beta )$ equals-

• A. $1$
• B. $2$
• C. $3$
• D. $None$ $of$ $these$

Answer 1

The correct option is A $1$

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

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

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

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

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

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