What is the value of this logarithmic/trigonometric expression? I need the value of this expression, when simplified.
$$ \log_{10}(\cot(1)°) + \log_{10}(\cot(2°)) + \cdots + \log_{10}(\cot(89°))$$
All the $\log$ have base $10$.
 A: The sum is essentially 
\begin{align}
S &= \sum_{k=1}^{89} \log_{10}\left(\cot\left(\frac{k \pi}{180}\right)\right) \\
&= \sum_{k=1}^{44} \log_{10}\left(\cot\left(\frac{k \pi}{180}\right)\right) + \log_{10}\left(\cot\left(\frac{\pi}{4}\right)\right) + \sum_{k=46}^{89} \log_{10}\left(\cot\left(\frac{k \pi}{180}\right)\right) \\
&= \sum_{k=1}^{44} \left[\log_{10}\left(\cot\left(\frac{k \pi}{180}\right)\right) + \log_{10}\left(\cot\left(\frac{\pi}{2} - \frac{k \pi}{180} \right)\right) \right] + \log_{10}\left(\cot\left(\frac{\pi}{4}\right)\right) \\
&= \sum_{k=1}^{44} \log_{10}\left(\cot\left(\frac{k \pi}{180}\right) \, \tan\left(\frac{k \pi}{180}\right)\right) + \log_{10}\left(\cot\left(\frac{\pi}{4}\right)\right) \\
&= \sum_{k=1}^{44} \log_{10}(1) + \log_{10}(1) \\ 
&= \sum_{k=1}^{45} \log_{10}(1) = 0. 
\end{align}
A: Another take on this is of course to show it graphically:


I find this helps to add some intuition to the excellent answers already posted.
A: $$\log(\cot\theta) + \log(\cot(90-\theta)) = \log(\cot\theta) + \log(\tan\theta) = \log(\cot\theta\tan\theta) = \log(1) = 0$$
