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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 $10$.

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    $\begingroup$ Hint: $\log (cot x^{\circ})+\log (\cot (90^{\circ}-x^{\circ}))=0.$ $\endgroup$ – Anurag A Jun 30 '17 at 3:56
  • $\begingroup$ @Anurag A How ? Please explain. $\endgroup$ – Ravi Prakash Jun 30 '17 at 3:58
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$$\log(\cot\theta) + \log(\cot(90-\theta)) = \log(\cot\theta) + \log(\tan\theta) = \log(\cot\theta\tan\theta) = \log(1) = 0$$

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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}

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    $\begingroup$ Fully explained, nice. But, You are finding the sum from 0° to 90° instead of 1° to 89° $\endgroup$ – Ravi Prakash Jun 30 '17 at 4:23
  • $\begingroup$ One more thing, could you write the answer without summations? I'll accept it! $\endgroup$ – Ravi Prakash Jun 30 '17 at 4:28
  • $\begingroup$ And why are you converting the degrees into radians? $\endgroup$ – Ravi Prakash Jun 30 '17 at 4:31
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    $\begingroup$ There is a mistake between 2nd and 3rd line because $k=45$ does not have a matching term in the second summation. Luckily it does not affect the final result because it is zero. $\endgroup$ – Rad80 Jun 30 '17 at 9:01
  • $\begingroup$ @RaviPrakash It's really from $1^\circ$ to $89^\circ$ already. At $0^\circ$ and $90^\circ$, the summand would be $\log_{10}(0)+\log_{10}(\infty)$. So we just need to adjust the claimed limits of summation to match the actual limits of summation. $\endgroup$ – Teepeemm Jun 30 '17 at 15:34
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Another take on this is of course to show it graphically:

log(cot(θ))

Running Sum of log(cot(θ))

I find this helps to add some intuition to the excellent answers already posted.

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