If `Delta = a^(2)-(b-c)^(2), Delta` is the area of the `Delta ABC` then `tan A =` ?

Answer

To solve the problem, we need to find the value of \( \tan A \) given that the area \( \Delta \) of triangle \( ABC \) is expressed as:

\[ \Delta = a^2 - (b - c)^2 \]

### Step-by-Step Solution:

1. **Express the Area in Terms of Sides**:
We know that the area \( \Delta \) of triangle \( ABC \) can also be expressed using the formula:
\[ \Delta = \frac{1}{2}ab \sin C \]
However, we will first manipulate the expression given in the problem.

2. **Rewrite the Expression**:
The expression for \( \Delta \) can be simplified:
\[ \Delta = a^2 - (b - c)^2 \]
This can be factored using the difference of squares:
\[ \Delta = (a + (b - c))(a - (b - c)) = (a + b - c)(a - b + c) \]

3. **Relate to Semi-Perimeter**:
In triangle geometry, we know that:
\[ s = \frac{a + b + c}{2} \]
Therefore, we can express \( a + b - c \) and \( a - b + c \) in terms of \( s \):
\[ a + b - c = (s + c) \quad \text{and} \quad a - b + c = (s + b) \]
So we can rewrite \( \Delta \):
\[ \Delta = (s + c)(s + b) \]

4. **Using the Area Formula**:
We know that the area can also be expressed in terms of the semi-perimeter \( s \):
\[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]
Equating the two expressions for \( \Delta \):
\[ (s + c)(s + b) = \sqrt{s(s-a)(s-b)(s-c)} \]

5. **Finding \( \tan A \)**:
We can use the relation:
\[ \tan \frac{A}{2} = \frac{(s-b)(s-c)}{s(s-a)} \]
Given that \( \Delta = \frac{1}{4} \), we can substitute this into the formula for \( \tan \frac{A}{2} \):
\[ \tan \frac{A}{2} = \frac{1}{4} \]

6. **Using the Double Angle Formula**:
We know that:
\[ \tan A = \frac{2 \tan \frac{A}{2}}{1 - \tan^2 \frac{A}{2}} \]
Substituting \( \tan \frac{A}{2} = \frac{1}{4} \):
\[ \tan A = \frac{2 \cdot \frac{1}{4}}{1 - \left(\frac{1}{4}\right)^2} = \frac{\frac{1}{2}}{1 - \frac{1}{16}} = \frac{\frac{1}{2}}{\frac{15}{16}} = \frac{1}{2} \cdot \frac{16}{15} = \frac{8}{15} \]

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