Question

If $\tan \alpha =\frac{x}{x+1}$ and $\tan \beta =\frac{1}{2x+1}$, then find $\alpha +\beta$.

Answer 1

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

It is known that $\tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$, then,

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

$=\frac{\frac{x\left(2x+1\right)+\left(x+1\right)}{\left(x+1\right)\left(2x+1\right)}}{\frac{\left(x+1\right)\left(2x+1\right)-x}{\left(x+1\right)\left(2x+1\right)}}$

$=\frac{2x^2+x+x+1}{2x^2+x+2x+1-x}$

$=\frac{2x^2+2x+1}{2x^2+2x+1}$

$=1$

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

$\alpha +\beta =\frac{\pi }{4}$