Instead of pairing all 89 terms, use a symmetry substitution on the sum itself. Let \(S\) denote the sum:
\[S = \sum_{k=1}^{89} \log \tan k^{\circ}\]
Replace \(k\) with \(90-k\) in the same sum (since the terms just run in reverse order, the sum is unchanged):
\[S = \sum_{k=1}^{89} \log \tan (90^{\circ}-k^{\circ}) = \sum_{k=1}^{89} \log \cot k^{\circ}\]
Adding the original sum and this rewritten version:
\[2S = \sum_{k=1}^{89} \left(\log \tan k^{\circ} + \log \cot k^{\circ}\right) = \sum_{k=1}^{89} \log(\tan k^{\circ}\cdot\cot k^{\circ}) = \sum_{k=1}^{89} \log 1 = 0\]
So \(2S=0\), which gives \(S=0\).
The value of the given expression is 0.
Hence, the correct answer is Option C: 0.