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

Options

A. `(4)/(15)`
B. `(8)/(15)`
C. `(11)/(15)`
D. `(13)/(15)`

Explanation

To find \( \tan A \) given that \( \Delta = a^2 - (b - c)^2 \) is the area of triangle \( ABC \), we can follow these steps:

### Step 1: Understand the area of the triangle
The area \( \Delta \) of triangle \( ABC \) can also be expressed using the formula:
\[ \Delta = \frac{1}{2}ab \sin A \]
where \( a \) is the length of side \( BC \), \( b \) is the length of side \( AC \), and \( A \) is the angle opposite side \( a \).

### Step 2: Rewrite the area expression
Given \( \Delta = a^2 - (b - c)^2 \), we can expand the right-hand side:
\[ \Delta = a^2 - (b^2 - 2bc + c^2) = a^2 - b^2 + 2bc - c^2 \]

### Step 3: Set the two expressions for area equal
Now we have two expressions for the area:
\[ \frac{1}{2}ab \sin A = a^2 - b^2 + 2bc - c^2 \]

### Step 4: Solve for \( \sin A \)
Rearranging gives:
\[ \sin A = \frac{2(a^2 - b^2 + 2bc - c^2)}{ab} \]

### Step 5: Use the relationship between \( \tan A \) and \( \sin A \)
We know that:
\[ \tan A = \frac{\sin A}{\cos A} \]
To find \( \tan A \), we need to express \( \cos A \) in terms of the sides of the triangle. We can use the cosine rule:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

### Step 6: Substitute \( \sin A \) and \( \cos A \) into \( \tan A \)
Substituting the expressions we have:
\[ \tan A = \frac{\frac{2(a^2 - b^2 + 2bc - c^2)}{ab}}{\frac{b^2 + c^2 - a^2}{2bc}} \]

### Step 7: Simplify the expression
This can be simplified further:
\[ \tan A = \frac{4(a^2 - b^2 + 2bc - c^2)bc}{ab(b^2 + c^2 - a^2)} \]

### Final Expression for \( \tan A \)
Thus, we have:
\[ \tan A = \frac{4bc(a^2 - b^2 + 2bc - c^2)}{ab(b^2 + c^2 - a^2)} \]