Let `f:[0,(pi)/(2)]toR` be continuous and satisfy `f'(x)=(1)/(1+cosx)` for all `x in(0,(pi)/(2))`. If f(0)=3 then `f((pi)/(2))` has the value equal to :

Answer

To solve the problem, we need to find the value of \( f\left(\frac{\pi}{2}\right) \) given that \( f'(x) = \frac{1}{1 + \cos x} \) and \( f(0) = 3 \).

### Step-by-Step Solution:

1. **Identify the given information:**
We have:
\[ f'(x) = \frac{1}{1 + \cos x} \]
and
\[ f(0) = 3. \]

2. **Rewrite the derivative using a trigonometric identity:**
We know that:
\[ 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right). \]
Therefore, we can rewrite \( f'(x) \):
\[ f'(x) = \frac{1}{2 \cos^2\left(\frac{x}{2}\right)} = \frac{1}{2} \sec^2\left(\frac{x}{2}\right). \]

3. **Integrate \( f'(x) \):**
Now, we integrate \( f'(x) \):
\[ f(x) = \int f'(x) \, dx = \int \frac{1}{2} \sec^2\left(\frac{x}{2}\right) \, dx. \]
The integral of \( \sec^2(u) \) is \( \tan(u) \), where \( u = \frac{x}{2} \). Thus,
\[ \int \sec^2\left(\frac{x}{2}\right) \, dx = 2 \tan\left(\frac{x}{2}\right) + C. \]
Therefore,
\[ f(x) = \frac{1}{2} \cdot 2 \tan\left(\frac{x}{2}\right) + C = \tan\left(\frac{x}{2}\right) + C. \]

4. **Use the initial condition to find \( C \):**
We know \( f(0) = 3 \):
\[ f(0) = \tan\left(0\right) + C = 0 + C = C. \]
Thus, \( C = 3 \).

5. **Write the final expression for \( f(x) \):**
Therefore, we have:
\[ f(x) = \tan\left(\frac{x}{2}\right) + 3. \]

6. **Evaluate \( f\left(\frac{\pi}{2}\right) \):**
Now, we calculate \( f\left(\frac{\pi}{2}\right) \):
\[ f\left(\frac{\pi}{2}\right) = \tan\left(\frac{\pi}{4}\right) + 3 = 1 + 3 = 4. \]

### Final Answer:
Thus, the value of \( f\left(\frac{\pi}{2}\right) \) is \( \boxed{4} \).