`logtan1^@+logtan2^@+.....+logtan89^@` is :

Options

A. 1
B. `1/sqrt2`
C. 0
D. -1

Answer

To solve the expression \( \log(\tan 1^\circ) + \log(\tan 2^\circ) + \ldots + \log(\tan 89^\circ) \), we can use properties of logarithms and the relationship between tangent and cotangent.

### Step-by-Step Solution:

1. **Use the property of logarithms**:
\[ \log(a) + \log(b) = \log(ab) \]
Therefore, we can combine the logs:
\[ \log(\tan 1^\circ \tan 2^\circ \tan 3^\circ \ldots \tan 89^\circ) \]

2. **Recognize the symmetry in tangent**:
We know that:
\[ \tan(90^\circ - x) = \cot(x) \]
Thus, we can pair the terms:
\[ \tan 1^\circ \tan 89^\circ = \tan 1^\circ \cot 1^\circ = 1 \]
\[ \tan 2^\circ \tan 88^\circ = \tan 2^\circ \cot 2^\circ = 1 \]
Continuing this way, we can see that:
\[ \tan k^\circ \tan(90^\circ - k^\circ) = 1 \quad \text{for } k = 1, 2, \ldots, 44 \]

3. **Count the pairs**:
There are 44 pairs from \( \tan 1^\circ \) to \( \tan 44^\circ \) and their corresponding cotangents. The middle term is \( \tan 45^\circ \), which equals 1.

4. **Combine the products**:
The product of all these terms can be expressed as:
\[ \tan 1^\circ \tan 89^\circ \times \tan 2^\circ \tan 88^\circ \times \ldots \times \tan 44^\circ \tan 46^\circ \times \tan 45^\circ \]
Since each pair equals 1, we have:
\[ \tan 1^\circ \tan 89^\circ = 1, \tan 2^\circ \tan 88^\circ = 1, \ldots, \tan 44^\circ \tan 46^\circ = 1 \]
and \( \tan 45^\circ = 1 \).

5. **Final calculation**:
Thus, the entire product simplifies to:
\[ \tan 1^\circ \tan 2^\circ \ldots \tan 89^\circ = 1^{44} \times 1 = 1 \]
Therefore:
\[ \log(\tan 1^\circ \tan 2^\circ \ldots \tan 89^\circ) = \log(1) = 0 \]

### Conclusion:
The value of \( \log(\tan 1^\circ) + \log(\tan 2^\circ) + \ldots + \log(\tan 89^\circ) \) is \( 0 \).