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# If $\tan\alpha=\large\frac{x}{x+1}$ and $\tan \beta=\large\frac{1}{2x+1}$ then $\alpha+\beta$ is

$(a)\;0\qquad(b)\;\large\frac{\pi}{4}$$\qquad(c)\;\large\frac{\pi}{3}$$\qquad(d)\;\large\frac{\pi}{2}$

We have $\tan(\alpha+\beta)=\large\frac{\tan\alpha+\tan\beta}{1-\tan \alpha\tan\beta}$
$\qquad\qquad\qquad\quad\;\;=\large\frac{\Large\frac{x}{x+1}+\frac{1}{2x+1}}{1-\Large\frac{x}{x+1}\frac{1}{2x+1}}$
$\qquad\qquad\qquad\quad\;\;=\large\frac{x(2x+1)+(x+1)}{(x+1)(2x+1)-x}$
$\qquad\qquad\qquad\quad\;\;=\large\frac{2x^2+2x+1}{2x^2+2x+1}$
$\qquad\qquad\qquad\quad\;\;=1$
$\Rightarrow \large\frac{\pi}{4}$
Hence $\alpha+\beta=\large\frac{\pi}{4}$