Question:
If $\tan A=\frac{a}{a+1}$ and $\tan B=\frac{1}{2 a+1}$, then the value of $A+B$ is
(a) 0
(b) $\frac{\pi}{2}$
(c) $\frac{\pi}{3}$
(d) $\frac{\pi}{4}$
Solution:
(d) $\frac{\pi}{4}$
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$
$=\frac{\frac{a}{a+1}+\frac{1}{2 a+1}}{1-\frac{a}{(a+1)(2 a+1)}}$
$=\frac{2 a^{2}+a+a+1}{2 a^{2}+3 a+1-a}$
$=\frac{2 a^{2}+2 a+1}{2 a^{2}+2 a+1}$
$=1$
Therefore, $A+B=\tan ^{-1}(1)=\frac{\pi}{4}$.
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