Arccotangent Sum

Using the arctan sum formula, we have that

$\cot^{-1}(7) + \cot^{-1}(13) = \tan^{-1}(\frac{1}{7}) + \tan^{-1}(\frac{1}{13}) = \tan^{-1}\left(\dfrac{\frac{1}{7} + \frac{1}{13}}{1 - \frac{1}{7}*\frac{1}{13}}\right) = \tan^{-1}(\frac{2}{9}).$

Similarly, $\cot^{-1}{(21)} + \cot^{-1}{(31)} = \tan^{-1}\left(\dfrac{\frac{1}{21} + \frac{1}{31}}{1 - \frac{1}{21}*\frac{1}{31}}\right) = \tan^{-1}(\frac{2}{25}).$

Since $\tan^{-1}(\frac{2}{9}) + \tan^{-1}(\frac{2}{25}) = \tan^{-1}\left(\dfrac{\frac{2}{9} + \frac{2}{25}}{1 - \frac{2}{9}*\frac{2}{25}}\right) = \tan^{-1}(\frac{68}{225}) = \tan^{-1}(\frac{4}{13}) = \cot^{-1}(\frac{13}{4}),$

the given expression comes out to $\cot(\cot^{-1}(\frac{13}{4})) = \dfrac{13}{4} = \boxed{3.25}.$

$S=\cot\left(\arctan\left(\dfrac{1}{7}\right)+\arctan\left(\dfrac{1}{13}\right)+\arctan\left(\dfrac{1}{21}\right)+\arctan\left(\dfrac{1}{31}\right)\right)$

Let $a=7+i$ , $b=13+i$ , $c=21+i$ and $d=31+i$ . Then:

$S=\cot(\arg(a)+\arg(b)+\arg(c)+\arg(d)))$

$S=\cot(\arg(abcd)$

$S=\cot(\arg((7+i)(13+i)(21+i)(31+i)))$

$S=\cot(\arg((10)(9+2i)(26)(25+2i)))$

We can ignore the $10$ and the $26$ , since they don't affect the argument:

$S=\cot(\arg(221+68i))$

$S=\dfrac{221}{68}=\boxed{\dfrac{13}{4}}$

I used similar methods as others here, but first derived $\cot \alpha = x, \cot\beta = y\ \Longrightarrow\ \cot(\alpha+\beta) = \frac{xy-1}{x+y};$ or, with fractions, $\cot \alpha = \frac x u, \cot\beta = \frac y v\ \Longrightarrow\ \cot(\alpha+\beta) = \frac{xy-uv}{xv+yu}.$

Applying this to the given values in pairs, $\cot(\cot^{-1}(7) + \cot^{-1}(13)) = \frac{7\cdot 13-1}{7+13} = \frac{90}{20} = \frac 9 2;$ $\cot(\cot^{-1}(21) + \cot^{-1}(31)) = \frac{21\cdot 31-1}{21+31} = \frac{650}{52} = \frac {25}2;$ $\cot\left(\cot^{-1}\left(\frac 92\right) +\cot^{-1}\left(\frac {25}2\right)\right) = \frac{9\cdot 25-2\cdot 2}{9\cdot 2+25\cdot 2} = \frac{221}{68} = \frac {13}4 = \boxed{3.25}.$