Discussion
\(\cot^{-1}(1) + \cot^{-1}(2) + \cot^{-1}(3) =\)
Answer: B
The sum is \(\frac{\pi}{2}\).
For positive arguments \(\cot^{-1}a = \tan^{-1}\frac{1}{a}\). Then
\[ \begin{aligned} \tan^{-1}\tfrac{1}{2} + \tan^{-1}\tfrac{1}{3} &= \tan^{-1}\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = \tan^{-1}1 = \tfrac{\pi}{4} \\ \text{sum} &= \tfrac{\pi}{4} + \tfrac{\pi}{4} = \tfrac{\pi}{2} \end{aligned} \]
\(\frac{\pi}{4}\) is only the first term or the last two terms; \(\pi\) and \(\frac{3\pi}{2}\) are too large for three angles each less than \(\frac{\pi}{4}\) or equal to it.
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