Question

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

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $0$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B

$\frac{\pi }{4}$

Explanation for the correct option:

Sum of trigonometric ratios:

$\tan \left(\alpha \right)=\frac{m}{m+1},\tan \left(\beta \right)=\frac{1}{2m+1}$

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

$\begin{array}{rcl}\tan \left(\alpha +\beta \right)& =& \frac{{\displaystyle \frac{m}{m+1}}+{\displaystyle \frac{1}{2m+1}}}{1-\left({\displaystyle \frac{m}{m+1}}\right)\left({\displaystyle \frac{1}{2m+1}}\right)}\\ & =& \frac{\left[{\displaystyle \frac{m\left(2m+1\right)+1\left(m+1\right)}{\left(m+1\right)\left(2m+1\right)}}\right]}{{\displaystyle \frac{\left(m+1\right)\left(2m+1\right)-m}{\left(m+1\right)\left(2m+1\right)}}}\\ & =& \frac{2m^2+m+m+1}{2m^2+m+2m+1-m}\\ & =& \frac{2m^2+2m+1}{2m^2+2m+1}\\ & =& 1\\ \tan \left(\alpha +\beta \right)& =& \tan \left(\frac{\pi }{4}\right)\left[\because tan\frac{\pi }{4}=1\right]\\ \therefore \left(\alpha +\beta \right)& =& \frac{\pi }{4}\end{array}$

Hence, the correct option is (B).