Question

If $\tan A=\frac{a}{a+1}\mathrm{and}\tan B=\frac{1}{2a+1}$, then the value of A + B is

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

Answer 1

(d) $\frac{\mathrm{\pi }}{4}$

$$\begin{gathered} \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \\ =\frac{{\displaystyle \frac{a}{a+1}}+{\displaystyle \frac{1}{2a+1}}}{1-{\displaystyle \frac{a}{\left(a+1\right)(2a+1)}}} \\ =\frac{2a^2+a+a+1}{2a^2+3a+1-a} \\ =\frac{2a^2+2a+1}{2a^2+2a+1} \\ =1 \\ \mathrm{Therefore},A+B={\tan }^{-1}\left(1\right)=\frac{\mathrm{\pi }}{4}. \end{gathered}$$