The function `f(x)=(1-sinx+cosx)/(1+sinx+cosx)` is nto defined at `x=pi`. The value of `f(pi)` so that f(x) is continuous at `x=pi` is:

Options

A. `-(1)/(2)`
B. `(1)/(2)`
C. `-1`
D. 1

Answer

To find the value of \( f(\pi) \) such that the function \( f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x} \) is continuous at \( x = \pi \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches \( \pi \).

### Step-by-step Solution:

1. **Evaluate \( f(\pi) \)**:
\[ f(\pi) = \frac{1 - \sin(\pi) + \cos(\pi)}{1 + \sin(\pi) + \cos(\pi)} \]
Since \( \sin(\pi) = 0 \) and \( \cos(\pi) = -1 \):
\[ f(\pi) = \frac{1 - 0 - 1}{1 + 0 - 1} = \frac{0}{0} \]
This is an indeterminate form, so we need to apply L'Hôpital's Rule.

2. **Differentiate the numerator and denominator**:
- The numerator is \( 1 - \sin x + \cos x \).
- The derivative of the numerator:
\[ \frac{d}{dx}(1 - \sin x + \cos x) = -\cos x - \sin x \]
- The denominator is \( 1 + \sin x + \cos x \).
- The derivative of the denominator:
\[ \frac{d}{dx}(1 + \sin x + \cos x) = \cos x + \sin x \]

3. **Apply L'Hôpital's Rule**:
We now take the limit as \( x \) approaches \( \pi \):
\[ \lim_{x \to \pi} \frac{-\cos x - \sin x}{\cos x + \sin x} \]

4. **Substitute \( x = \pi \)**:
\[ \text{Numerator: } -\cos(\pi) - \sin(\pi) = -(-1) - 0 = 1 \]
\[ \text{Denominator: } \cos(\pi) + \sin(\pi) = -1 + 0 = -1 \]
Therefore, the limit is:
\[ \lim_{x \to \pi} f(x) = \frac{1}{-1} = -1 \]

5. **Conclusion**:
To make \( f(x) \) continuous at \( x = \pi \), we set:
\[ f(\pi) = -1 \]

### Final Answer:
The value of \( f(\pi) \) so that \( f(x) \) is continuous at \( x = \pi \) is \( -1 \).