If `a_(1),a_(2),a_(3),……a_(87),a_(88),a_(89)` are the arithmetic means between `1` and `89`, then `sum_(r=1)^(89)log(tan(a_(r ))^(@))` is equal to

Options

A. `0`
B. `1`
C. `log_(2)3`
D. `log5`

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript and break it down into clear steps.

### Step-by-Step Solution:

1. **Understanding the Arithmetic Means**:
We have \( a_1, a_2, a_3, \ldots, a_{89} \) as the arithmetic means between 1 and 89. This means:
\[ a_r = 1 + \frac{(r)(89 - 1)}{90} \quad \text{for } r = 1, 2, \ldots, 89 \]

2. **Expression for the Summation**:
We need to evaluate:
\[ \sum_{r=1}^{89} \log(\tan(a_r)) \]
This can be rewritten using the property of logarithms:
\[ = \log\left(\prod_{r=1}^{89} \tan(a_r)\right) \]

3. **Using the Property of Arithmetic Means**:
Since \( a_r \) are arithmetic means, we know that:
\[ a_r + a_{90 - r} = 90 \quad \text{for } r = 1, 2, \ldots, 44 \]
This implies:
\[ a_{90 - r} = 90 - a_r \]

4. **Using the Tangent Identity**:
From the tangent identity:
\[ \tan(90 - x) = \cot(x) \]
We can express:
\[ \tan(a_{90 - r}) = \tan(90 - a_r) = \cot(a_r) \]

5. **Pairing the Terms**:
The product of terms can be paired:
\[ \tan(a_r) \tan(a_{90 - r}) = \tan(a_r) \cot(a_r) = 1 \]
Thus, for \( r = 1 \) to \( 44 \):
\[ \prod_{r=1}^{44} \tan(a_r) \tan(a_{90 - r}) = 1 \]

6. **Considering the Middle Term**:
The middle term \( a_{45} \) is:
\[ a_{45} + a_{45} = 90 \implies 2a_{45} = 90 \implies a_{45} = 45 \]
Therefore:
\[ \tan(a_{45}) = \tan(45) = 1 \]

7. **Final Calculation**:
Now substituting back into the logarithmic expression:
\[ \prod_{r=1}^{89} \tan(a_r) = 1 \quad \text{(from pairs)} \quad \text{and} \quad \tan(a_{45}) = 1 \]
Thus:
\[ \log\left(\prod_{r=1}^{89} \tan(a_r)\right) = \log(1) = 0 \]

### Conclusion:
The value of the summation \( \sum_{r=1}^{89} \log(\tan(a_r)) \) is:
\[ \boxed{0} \]