`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 \( \sum_{k=1}^{89} \log_e \tan\left(\frac{\pi k}{180}\right) \), we can follow these steps:

### Step 1: Define the Summation
Let:
\[ y = \sum_{k=1}^{89} \log_e \tan\left(\frac{\pi k}{180}\right) \]

### Step 2: Use Properties of Logarithms
We can use the property of logarithms that states \( \log(a) + \log(b) = \log(ab) \). Thus, we can rewrite \( y \) as:
\[ y = \log_e \left( \tan\left(\frac{\pi}{180}\right) \cdot \tan\left(\frac{2\pi}{180}\right) \cdots \tan\left(\frac{89\pi}{180}\right) \right) \]

### Step 3: Pairing the Terms
Notice that \( \tan\left(\frac{(90-k)\pi}{180}\right) = \cot\left(\frac{k\pi}{180}\right) \). Therefore, we can pair the terms:
- For \( k = 1 \), we have \( \tan\left(\frac{\pi}{180}\right) \) and \( \tan\left(\frac{89\pi}{180}\right) = \cot\left(\frac{\pi}{180}\right) \).
- For \( k = 2 \), we have \( \tan\left(\frac{2\pi}{180}\right) \) and \( \tan\left(\frac{88\pi}{180}\right) = \cot\left(\frac{2\pi}{180}\right) \).
- This continues up to \( k = 44 \) and \( k = 45 \).

### Step 4: Simplifying the Product
Each pair \( \tan\left(\frac{k\pi}{180}\right) \cdot \cot\left(\frac{k\pi}{180}\right) = 1 \). Therefore, we can write:
\[ y = \log_e \left( \prod_{k=1}^{44} 1 \cdot \tan\left(\frac{45\pi}{180}\right) \right) \]
Since \( \tan\left(\frac{45\pi}{180}\right) = 1 \), we have:
\[ y = \log_e(1) = 0 \]

### Step 5: Conclusion
Thus, the final result is:
\[ \sum_{k=1}^{89} \log_e \tan\left(\frac{\pi k}{180}\right) = 0 \]