`(dy)/(dx) ((1+ cos x)/(y)) =- sin x and f ((pi)/(2)) =-1,` then `f (0)` is:

Options

A. -2
B. -1
C. -3
D. -4

Answer

To solve the differential equation given by

\[ \frac{dy}{dx} \cdot \frac{1 + \cos x}{y} = -\sin x \]

with the initial condition \( f\left(\frac{\pi}{2}\right) = -1 \), we will follow these steps:

### Step 1: Separate the Variables
We can rearrange the equation to separate the variables \( y \) and \( x \):

\[ \frac{1}{y} dy = -\frac{\sin x}{1 + \cos x} dx \]

### Step 2: Integrate Both Sides
Now we will integrate both sides:

\[ \int \frac{1}{y} dy = \int -\frac{\sin x}{1 + \cos x} dx \]

The left-hand side integrates to:

\[ \ln |y| + C_1 \]

For the right-hand side, we can simplify the integral. Notice that:

\[ -\frac{\sin x}{1 + \cos x} = -\frac{d(1 + \cos x)}{1 + \cos x} \]

Thus, we can perform the substitution \( t = 1 + \cos x \), which gives us:

\[ \int -\frac{1}{t} dt = -\ln |t| + C_2 \]

Substituting back for \( t \):

\[ -\ln |1 + \cos x| + C_2 \]

### Step 3: Combine the Results
Combining both integrals, we have:

\[ \ln |y| = -\ln |1 + \cos x| + C \]

where \( C = C_2 - C_1 \).

### Step 4: Exponentiate to Solve for \( y \)
Exponentiating both sides gives:

\[ |y| = e^{C} \cdot \frac{1}{|1 + \cos x|} \]

Let \( C' = e^{C} \), we can write:

\[ y = \frac{C'}{1 + \cos x} \]

### Step 5: Apply Initial Condition
Now we apply the initial condition \( f\left(\frac{\pi}{2}\right) = -1 \):

At \( x = \frac{\pi}{2} \):

\[ y = \frac{C'}{1 + \cos\left(\frac{\pi}{2}\right)} = \frac{C'}{1 + 0} = C' \]

Setting this equal to -1 gives:

\[ C' = -1 \]

### Step 6: Write the Final Solution
Thus, the solution for \( y \) becomes:

\[ y = \frac{-1}{1 + \cos x} \]

### Step 7: Find \( f(0) \)
Now we need to find \( f(0) \):

\[ f(0) = y(0) = \frac{-1}{1 + \cos(0)} = \frac{-1}{1 + 1} = \frac{-1}{2} \]

### Final Answer
Thus, the value of \( f(0) \) is:

\[ \boxed{-\frac{1}{2}} \]