The problem asks us to evaluate the sum of the logarithmic series: \( \log \tan 1^\circ + \log \tan 2^\circ + \dots + \log \tan 89^\circ \).

Logarithm Properties: Sum to Product

First, let's recall a fundamental property of logarithms that allows us to combine a sum of logarithms into a single logarithm of a product. This property is:

• For any positive numbers \(A\) and \(B\), and a valid base, \( \log A + \log B = \log (A \cdot B) \).

Applying this property repeatedly to the given series, we can express the entire sum as a single logarithm of a product of all the tangent terms:

\[ \text{Sum} = \log (\tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdot \dots \cdot \tan 88^\circ \cdot \tan 89^\circ) \]

Trigonometric Identities: Complementary Angles

Next, we need to simplify the product of tangent terms inside the logarithm. This requires the use of a key trigonometric identity involving complementary angles:

• The tangent of an angle \( (90^\circ - \theta) \) is equal to the cotangent of \( \theta \).
  Symbolically, \( \tan (90^\circ - \theta) = \cot \theta \).
• We also know that the product of tangent and cotangent of the same angle is 1.
  That is, \( \tan \theta \cdot \cot \theta = 1 \). This is because \( \cot \theta = \frac{1}{\tan \theta} \).

Simplifying the Tangent Product

Let's consider the product of tangents: \( P = \tan 1^\circ \cdot \tan 2^\circ \cdot \dots \cdot \tan 89^\circ \). We can pair terms from the beginning and the end of the product. This strategy simplifies the expression significantly:

• The first term is \( \tan 1^\circ \).
• The last term is \( \tan 89^\circ \). Using the identity \( \tan (90^\circ - \theta) = \cot \theta \), we get \( \tan 89^\circ = \tan (90^\circ - 1^\circ) = \cot 1^\circ \).
• So, the product of the first and last terms is \( \tan 1^\circ \cdot \tan 89^\circ = \tan 1^\circ \cdot \cot 1^\circ \). As per our identity, this product equals 1.

This pairing pattern holds true for other terms as well:

• For the second term and second-to-last term: \( \tan 2^\circ \cdot \tan 88^\circ = \tan 2^\circ \cdot \tan (90^\circ - 2^\circ) = \tan 2^\circ \cdot \cot 2^\circ = 1 \).

This pairing continues. The series contains 89 terms. When we pair them up (e.g., \( (1^\circ, 89^\circ), (2^\circ, 88^\circ), \dots \)), there will be a middle term that is not paired. The middle term will be at the \( \frac{89+1}{2} = 45^\text{th} \) position, which is \( \tan 45^\circ \).

So, the entire product can be written as:

\[ P = (\tan 1^\circ \cdot \tan 89^\circ) \cdot (\tan 2^\circ \cdot \tan 88^\circ) \cdot \dots \cdot (\tan 44^\circ \cdot \tan 46^\circ) \cdot \tan 45^\circ \]

Since each pair \( (\tan x^\circ \cdot \tan (90^\circ - x^\circ)) \) simplifies to 1, the product becomes:

\[ P = (1) \cdot (1) \cdot \dots \cdot (1) \cdot \tan 45^\circ \]

We know that the value of \( \tan 45^\circ \) is 1. Therefore, the entire product \( P \) simplifies to:

\[ P = 1 \]

Final Logarithm Evaluation

Now, we substitute this simplified product back into our original logarithmic expression:

\[ \text{Sum} = \log (P) = \log (1) \]

Finally, we recall another fundamental property of logarithms:

• The logarithm of 1 to any valid base is always 0.
  That is, \( \log_b 1 = 0 \) for any base \( b > 0, b \neq 1 \).

Therefore, \( \log 1 = 0 \).

Conclusion: Series Sum

The sum of the series \( \log \tan 1^\circ + \log \tan 2^\circ + \dots + \log \tan 89^\circ \) is 0.