Question

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

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

Answer 1

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

Now,

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

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

Hence, the correct answer is option D.