Question

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

(a) $\frac{\mathrm{\pi }}{2}$ (b) $\frac{\mathrm{\pi }}{3}$ (c) $\frac{\mathrm{\pi }}{6}$ (d) $\frac{\mathrm{\pi }}{4}$

Answer 1

It is given that $\tan \alpha =\frac{x}{x+1}$ and $\tan \beta =\frac{1}{2x+1}$.

$$\begin{gathered} \tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } \\ =\frac{{\displaystyle \frac{x}{x+1}}+{\displaystyle \frac{1}{2x+1}}}{1-{\displaystyle \frac{x}{x+1}}\times {\displaystyle \frac{1}{2x+1}}} \\ =\frac{{\displaystyle \frac{x\left(2x+1\right)+\left(x+1\right)}{\left(x+1\right)\left(2x+1\right)}}}{{\displaystyle \frac{{\displaystyle \left(x+1\right)\left(2x+1\right)-x}}{{\displaystyle \left(x+1\right)\left(2x+1\right)}}}} \\ =\frac{{\displaystyle 2x^2+x+x+1}}{{\displaystyle 2x^2+3x+1-x}} \end{gathered}$$
$$\begin{gathered} =\frac{2x^2+2x+1}{2x^2+2x+1} \\ =1 \end{gathered}$$

$\therefore \alpha +\beta =\frac{\mathrm{\pi }}{4}\left(\tan \frac{\mathrm{\pi }}{4}=1\right)$

Hence, the correct answer is option D.