Discussion
If \(\tan\alpha = \frac{5}{6}\) and \(\tan\beta = \frac{1}{11}\) , where \(0 \lt \alpha, \beta \lt \frac{\pi}{2}\) then \(\alpha + \beta =\)
Answer: D
\(\alpha + \beta = \frac{\pi}{4}\).
Use the addition formula for tangent:
\[ \begin{aligned} \tan(\alpha + \beta) &= \frac{\frac{5}{6} + \frac{1}{11}}{1 - \frac{5}{6}\cdot\frac{1}{11}} \\ &= \frac{61/66}{61/66} = 1 \end{aligned} \]
Both angles are acute with small tangents, so \(0 \lt \alpha + \beta \lt \pi\) and \(\tan(\alpha + \beta) = 1\) gives \(\alpha + \beta = \frac{\pi}{4}\). \(\frac{\pi}{3}\) and \(\frac{\pi}{6}\) would need tangents \(\sqrt{3}\) and \(\frac{1}{\sqrt{3}}\).
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