Question

If $\tan \alpha =\frac{m}{m+1}$ and $\tan \beta =\frac{1}{2m+1}$ find the possible values of $(\alpha +\beta )$

• A. ${30}^{\circ }$
• B. ${90}^{\circ }$
• C. ${60}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is D ${45}^{\circ }$

Given that $\tan \alpha =\frac{m}{m+1}$ ……. (1) and $\tan \beta =\frac{1}{2m+1}$ …….. (2)

Find $\alpha +\beta =?$

We know that,

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

Put the value of $\tan \alpha$ and $\tan \beta$

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

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

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

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

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

$\Rightarrow \tan (\alpha +\beta )=\tan {45}^0$

$\Rightarrow \alpha +\beta ={45}^0$

Hence, this is the answer