`sum _(k =1) ^(89) log _(e) tan ((pik)/(180)) ` is equal to

Options

A. zero
B. 1
C. `log_(e) (89)`
D. none of these

Answer

To solve the problem, we need to evaluate the summation:

\[ y = \sum_{k=1}^{89} \log_e \tan\left(\frac{\pi k}{180}\right) \]

### Step 1: Rewrite the Summation
We can express the summation as follows:

\[ y = \log_e \left( \tan\left(\frac{\pi}{180}\right) \cdot \tan\left(\frac{2\pi}{180}\right) \cdots \tan\left(\frac{89\pi}{180}\right) \right) \]

### Step 2: Use the Property of Tangent
We know that:

\[ \tan\left(\frac{\pi k}{180}\right) = \cot\left(\frac{\pi (90-k)}{180}\right) \]

This means that:

\[ \tan\left(\frac{\pi k}{180}\right) \cdot \tan\left(\frac{\pi (90-k)}{180}\right) = 1 \]

### Step 3: Pairing Terms
We can pair the terms in the summation:

- For \( k = 1 \) and \( k = 89 \)
- For \( k = 2 \) and \( k = 88 \)
- ...
- For \( k = 44 \) and \( k = 46 \)

The middle term when \( k = 45 \) does not have a pair, and it is:

\[ \tan\left(\frac{45\pi}{180}\right) = 1 \]

### Step 4: Simplifying the Product
Since each pair of terms \( \tan\left(\frac{\pi k}{180}\right) \cdot \tan\left(\frac{\pi (90-k)}{180}\right) = 1 \), we can conclude that:

\[ \tan\left(\frac{\pi k}{180}\right) \cdot \tan\left(\frac{\pi (90-k)}{180}\right) = 1 \quad \text{for } k = 1, 2, \ldots, 44 \]

Thus, the product of all pairs contributes nothing to the logarithm, and we are left with:

\[ y = \log_e(1) = 0 \]

### Final Answer
Therefore, the value of the summation is:

\[ \boxed{0} \]