If `lim_(xto0) (p sin2x+(1-cos2x))/(x+tanx)=1`, then the value of `p` is___________.

Explanation

To solve the limit problem given by the expression:

\[ \lim_{x \to 0} \frac{p \sin 2x + (1 - \cos 2x)}{x + \tan x} = 1 \]

we will find the value of \( p \).

### Step-by-Step Solution:

1. **Identify the limit expression**:
We have:
\[ \lim_{x \to 0} \frac{p \sin 2x + (1 - \cos 2x)}{x + \tan x} \]

2. **Rewrite the denominator**:
We know that as \( x \to 0 \), \( \tan x \) can be approximated by \( x \). Thus, we can write:
\[ x + \tan x \approx x + x = 2x \]

3. **Multiply and divide by \( x \)**:
To simplify the limit, we multiply and divide the numerator by \( x \):
\[ \lim_{x \to 0} \frac{p \sin 2x + (1 - \cos 2x)}{x + \tan x} = \lim_{x \to 0} \frac{x(p \frac{\sin 2x}{x} + \frac{1 - \cos 2x}{x})}{x + \tan x} \]

4. **Factor out \( x \)**:
This gives us:
\[ = \lim_{x \to 0} \frac{x \left( p \frac{\sin 2x}{x} + \frac{1 - \cos 2x}{x} \right)}{x(1 + \frac{\tan x}{x})} \]
Cancelling \( x \) from numerator and denominator, we have:
\[ = \lim_{x \to 0} \frac{p \frac{\sin 2x}{x} + \frac{1 - \cos 2x}{x}}{1 + \frac{\tan x}{x}} \]

5. **Apply the limits**:
We know:
- \( \lim_{x \to 0} \frac{\sin 2x}{x} = 2 \)
- \( \lim_{x \to 0} \frac{1 - \cos 2x}{x} = 0 \) (using L'Hôpital's Rule or Taylor series)
- \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \)

Substituting these limits:
\[ = \frac{p \cdot 2 + 0}{1 + 1} = \frac{2p}{2} = p \]

6. **Set the limit equal to 1**:
Since we need this limit to equal 1:
\[ p = 1 \]

### Conclusion:
Thus, the value of \( p \) is:
\[ \boxed{1} \]