For `0leP,Qlepi/2`, if `sinP+cosQ=2`, then the value of `tan((P+Q)/(2))` is equal to

Options

A. 1
B. `(1)/(sqrt(2))`
C. `1/2`
D. `(sqrt(3))/(2)`

Answer

To solve the problem, we need to find the value of \( \tan\left(\frac{P+Q}{2}\right) \) given that \( \sin P + \cos Q = 2 \) for \( 0 \leq P, Q \leq \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Understanding the Equation**:
We have the equation:
\[ \sin P + \cos Q = 2 \]
The maximum value of \( \sin P \) is 1 (which occurs when \( P = \frac{\pi}{2} \)) and the maximum value of \( \cos Q \) is also 1 (which occurs when \( Q = 0 \)). Therefore, for the sum to equal 2, both \( \sin P \) and \( \cos Q \) must reach their maximum values.

2. **Setting the Values**:
From the above observation:
\[ \sin P = 1 \quad \text{and} \quad \cos Q = 1 \]
This implies:
\[ P = \frac{\pi}{2} \quad \text{and} \quad Q = 0 \]

3. **Calculating \( P + Q \)**:
Now, we calculate \( P + Q \):
\[ P + Q = \frac{\pi}{2} + 0 = \frac{\pi}{2} \]

4. **Finding \( \frac{P + Q}{2} \)**:
We need to find:
\[ \frac{P + Q}{2} = \frac{\frac{\pi}{2}}{2} = \frac{\pi}{4} \]

5. **Calculating \( \tan\left(\frac{P + Q}{2}\right) \)**:
Now we can find the tangent:
\[ \tan\left(\frac{P + Q}{2}\right) = \tan\left(\frac{\pi}{4}\right) \]
We know that:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]

### Final Answer:
Thus, the value of \( \tan\left(\frac{P + Q}{2}\right) \) is:
\[ \boxed{1} \]