Question

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

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{6}$
• D. $\frac{\pi}{4}$

Answer 1

Answer: (D)

Given $tan\alpha=\frac{m}{m+1}$,
$tan\beta=\frac{1}{2m+1}$
$\therefore tan\left(\alpha+\beta\right)=\frac{tan\,\alpha+tan\,\beta}{1-tan\,\alpha\,tan\,\beta}$
$=\frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1-\frac{m}{m+1}\times\frac{2}{2m+1}}$
$=\frac{2m^{2}+m+m+1}{\left(m+1\right)\left(2m+1\right)-m}$
$=\frac{2m^{2}+2m+1}{2m^{2}+2m+1}=1$
$\Rightarrow tan\left(\alpha+\beta\right)=1$
$\Rightarrow \alpha+\beta=\frac{\pi}{4}$