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Question
If \[\tan A = \frac{a}{a + 1}\text{ and } \tan B = \frac{1}{2a + 1}\]
Options
(a) 0
(b)\[\frac{\pi}{2}\]
(c) \[\frac{\pi}{3}\]
(d) \[\frac{\pi}{4}\]
Solution
(d)\[\frac{\pi}{4}\]
\[\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B} \]
\[ = \frac{\frac{a}{a + 1} + \frac{1}{2a + 1}}{1 - \frac{a}{\left( a + 1 \right)(2a + 1)}}\]
\[ = \frac{2 a^2 + a + a + 1}{2 a^2 + 3a + 1 - a}\]
\[ = \frac{2 a^2 + 2a + 1}{2 a^2 + 2a + 1}\]
\[ = 1\]
\[\text{ Therefore }, A + B = \tan^{- 1} (1) = \frac{\pi}{4} . \]
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