The function `f(x)=log((1+sinx)/(1-sinx))` is

Options

A. even
B. odd
C. neither even nor odd
D. both even and odd

Explanation

To determine whether the function \( f(x) = \log\left(\frac{1 + \sin x}{1 - \sin x}\right) \) is even, odd, or neither, we will follow these steps:

### Step 1: Find \( f(-x) \)
To check if the function is even or odd, we first need to calculate \( f(-x) \):
\[ f(-x) = \log\left(\frac{1 + \sin(-x)}{1 - \sin(-x)}\right) \]
Using the property of sine, we know that \( \sin(-x) = -\sin x \). Therefore, we can substitute:
\[ f(-x) = \log\left(\frac{1 - \sin x}{1 + \sin x}\right) \]

### Step 2: Simplify \( f(-x) \)
Next, we can rewrite \( f(-x) \):
\[ f(-x) = \log\left(\frac{1 - \sin x}{1 + \sin x}\right) \]
This can be expressed as:
\[ f(-x) = \log\left(\frac{1 + \sin x}{1 - \sin x}\right)^{-1} \]
Using the logarithmic property \( \log(a^{-1}) = -\log(a) \), we have:
\[ f(-x) = -\log\left(\frac{1 + \sin x}{1 - \sin x}\right) \]
Thus, we can write:
\[ f(-x) = -f(x) \]

### Step 3: Determine if \( f(x) \) is odd
Since we have established that:
\[ f(-x) = -f(x) \]
This satisfies the condition for the function to be classified as an odd function.

### Conclusion
Therefore, the function \( f(x) = \log\left(\frac{1 + \sin x}{1 - \sin x}\right) \) is an **odd function**.

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