If `[{:(,e^(x),sin x),(,cos x,ln(1+x))]:}=A+Bx+Cx^(2)+.......` then find the value of A and B.

Answer

To solve the given problem, we need to find the values of A and B from the expression given by the determinant of the matrix:

\[ \begin{vmatrix} e^x & \sin x \\ \cos x & \ln(1+x) \end{vmatrix} \]

This determinant can be calculated as follows:

### Step 1: Calculate the Determinant

The determinant of a 2x2 matrix is given by:

\[ \text{det}(A) = ad - bc \]

For our matrix, we have:

- \( a = e^x \)
- \( b = \sin x \)
- \( c = \cos x \)
- \( d = \ln(1+x) \)

Thus, the determinant is:

\[ \text{det}(A) = e^x \cdot \ln(1+x) - \sin x \cdot \cos x \]

### Step 2: Expand the Terms

Next, we will expand \( e^x \) and \( \ln(1+x) \) using their Taylor series expansions around \( x = 0 \):

1. **Expansion of \( e^x \)**:
\[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots \]

2. **Expansion of \( \ln(1+x) \)**:
\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \]

### Step 3: Substitute the Expansions

Now, substituting these expansions into the determinant:

\[ \text{det}(A) = (1 + x + \frac{x^2}{2} + \ldots)(x - \frac{x^2}{2} + \ldots) - \sin x \cdot \cos x \]

### Step 4: Expand the Product

Now, we will expand the product:

\[ (1 + x + \frac{x^2}{2})(x - \frac{x^2}{2}) = x + x^2 - \frac{x^3}{2} + \frac{x^2}{2} - \frac{x^3}{4} + \ldots \]

Combining like terms, we will focus on the coefficients of \( x \) and \( x^2 \):

- The coefficient of \( x \) is \( 1 \).
- The coefficient of \( x^2 \) is \( 1 - \frac{1}{2} = \frac{1}{2} \).

### Step 5: Calculate \( \sin x \cdot \cos x \)

Using the expansion for \( \sin x \) and \( \cos x \):

\[ \sin x = x - \frac{x^3}{6} + \ldots \]
\[ \cos x = 1 - \frac{x^2}{2} + \ldots \]

Thus,

\[ \sin x \cdot \cos x = (x - \frac{x^3}{6})(1 - \frac{x^2}{2}) = x - \frac{x^3}{6} - \frac{x^3}{2} + \ldots \]

The coefficient of \( x^2 \) from this product is \( 0 \).

### Step 6: Combine Results

Now we can combine our results:

\[ \text{det}(A) = (1)(x) + \left(\frac{1}{2} - 0\right)x^2 + \ldots \]

### Conclusion

From the expression \( \text{det}(A) = A + Bx + Cx^2 + \ldots \):

- The constant term \( A = 0 \).
- The coefficient of \( x \), \( B = 0 \).

Thus, the final values are:

\[ \boxed{A = 0, B = 0} \]