The area of a `DeltaABC` is `b^(2)-(c-a)^(2)`. Then ,tan B =

Options

A. `4/3`
B. `3/4`
C. 8/15
D. none of these

Answer

To solve the problem, we need to find \( \tan B \) given that the area of triangle \( ABC \) is \( \Delta = b^2 - (c - a)^2 \).

### Step-by-Step Solution:

1. **Identify the Area Expression**:
We start with the area of the triangle given as:
\[ \Delta = b^2 - (c - a)^2 \]

2. **Recognize the Difference of Squares**:
The expression \( b^2 - (c - a)^2 \) can be factored using the difference of squares formula:
\[ \Delta = (b + (c - a))(b - (c - a)) \]
This simplifies to:
\[ \Delta = (b + c - a)(b - c + a) \]

3. **Express in Terms of Semi-Perimeter**:
The semi-perimeter \( s \) of triangle \( ABC \) is given by:
\[ s = \frac{a + b + c}{2} \]
Therefore, we can express:
\[ b + c - a = 2s - 2a \]
and
\[ b - c + a = 2s - 2c \]
Thus, we can rewrite the area as:
\[ \Delta = (2s - 2a)(2s - 2c) = 4(s - a)(s - c) \]

4. **Relate Area to \( \tan B \)**:
We know from the triangle area formula that:
\[ \Delta = \frac{1}{2}ab \sin C \]
We also have the formula for \( \tan \frac{B}{2} \):
\[ \tan \frac{B}{2} = \frac{s - a}{s - c} \]
From our area expression, we can equate:
\[ 4(s - a)(s - c) = \Delta \]

5. **Using the Half-Angle Tangent Formula**:
We can use the identity:
\[ \tan B = \frac{2 \tan \frac{B}{2}}{1 - \tan^2 \frac{B}{2}} \]
Substituting \( \tan \frac{B}{2} \):
\[ \tan B = \frac{2 \cdot \frac{s - a}{s - c}}{1 - \left(\frac{s - a}{s - c}\right)^2} \]

6. **Substituting Values**:
We know from our previous steps that \( \Delta = 4(s - a)(s - c) \). Thus:
\[ \tan \frac{B}{2} = \frac{1}{4} \]
Therefore:
\[ \tan B = \frac{2 \cdot \frac{1}{4}}{1 - \left(\frac{1}{4}\right)^2} = \frac{\frac{1}{2}}{1 - \frac{1}{16}} = \frac{\frac{1}{2}}{\frac{15}{16}} = \frac{1}{2} \cdot \frac{16}{15} = \frac{8}{15} \]

### Final Answer:
Thus, the value of \( \tan B \) is:
\[ \tan B = \frac{8}{15} \]