Using the property $\log a + \log b = \log(ab)$:
$$\log(\tan 1^\circ) + \log(\tan 2^\circ) + \dots + \log(\tan 89^\circ) = \log(\tan 1^\circ \cdot \tan 2^\circ \cdot \dots \cdot \tan 89^\circ)$$
Using $\tan(90^\circ - \theta) = \cot \theta = \frac{1}{\tan \theta}$:
- $\tan 89^\circ = \cot 1^\circ = \frac{1}{\tan 1^\circ}$
- $\tan 88^\circ = \cot 2^\circ = \frac{1}{\tan 2^\circ}$
Pairing the complementary terms:
$$= \log\left( \left(\tan 1^\circ \cdot \frac{1}{\tan 1^\circ}\right) \left(\tan 2^\circ \cdot \frac{1}{\tan 2^\circ}\right) \dots \tan 45^\circ \right)$$
$$= \log(1) = 0$$
Correct Option: C
