Find the value of: `log(tan1^@)+log(tan2^@)+...+log(tan89^@).`

Answer

To find the value of \( \log(\tan 1^\circ) + \log(\tan 2^\circ) + \ldots + \log(\tan 89^\circ) \), we can follow these steps:

### Step 1: Use the property of logarithms
We know that \( \log A + \log B = \log(AB) \). Therefore, we can combine the logarithms:

\[ \log(\tan 1^\circ) + \log(\tan 2^\circ) + \ldots + \log(\tan 89^\circ) = \log(\tan 1^\circ \tan 2^\circ \tan 3^\circ \ldots \tan 89^\circ) \]

### Step 2: Analyze the product
Now, we need to analyze the product \( \tan 1^\circ \tan 2^\circ \tan 3^\circ \ldots \tan 89^\circ \). Notice that:

\[ \tan(90^\circ - x) = \cot x \]

This means:

\[ \tan 89^\circ = \cot 1^\circ, \quad \tan 88^\circ = \cot 2^\circ, \quad \ldots, \quad \tan 46^\circ = \cot 44^\circ \]

### Step 3: Pair the terms
We can pair the terms in the product:

\[ \tan 1^\circ \tan 89^\circ = \tan 1^\circ \cot 1^\circ = 1 \]
\[ \tan 2^\circ \tan 88^\circ = \tan 2^\circ \cot 2^\circ = 1 \]
\[ \vdots \]
\[ \tan 44^\circ \tan 46^\circ = \tan 44^\circ \cot 44^\circ = 1 \]

The middle term, \( \tan 45^\circ \), equals 1.

### Step 4: Calculate the total product
Since we have 44 pairs of terms that each multiply to 1, and the middle term \( \tan 45^\circ = 1 \), the entire product is:

\[ \tan 1^\circ \tan 2^\circ \ldots \tan 89^\circ = 1 \times 1 \times \ldots \times 1 = 1 \]

### Step 5: Find the logarithm
Now substituting back into our logarithm:

\[ \log(\tan 1^\circ \tan 2^\circ \ldots \tan 89^\circ) = \log(1) \]

### Step 6: Conclusion
Since \( \log(1) = 0 \), we conclude that:

\[ \log(\tan 1^\circ) + \log(\tan 2^\circ) + \ldots + \log(\tan 89^\circ) = 0 \]

### Final Answer
Thus, the value is:

\[ \boxed{0} \]