Step 1: Understanding the Concept:
The area of a triangle \( (\Delta) \) can be calculated using several formulas depending on the given data.
One common formula is \( \Delta = \frac{1}{2} ac \sin B \).
In this problem, a geometric expression for the area is given: \( \Delta = b^2 - (c - a)^2 \).
We need to manipulate this expression using the Law of Cosines and trigonometric identities to isolate \( \tan B \).
The cosine rule allows us to relate the side \( b \) with the sides \( a, c \) and angle \( B \).
Step 2: Key Formula or Approach:
1. Given Area \( \Delta = b^2 - (c - a)^2 \).
2. Standard Area \( \Delta = \frac{1}{2} ac \sin B \).
3. Cosine Rule: \( b^2 = a^2 + c^2 - 2ac \cos B \).
Step 3: Detailed Explanation:
Expand the given area expression:
\[ \Delta = b^2 - (c^2 + a^2 - 2ac) \]
\[ \Delta = b^2 - c^2 - a^2 + 2ac \]
Now, use the cosine rule for \( b^2 \):
\[ b^2 - a^2 - c^2 = -2ac \cos B \]
Substitute this into the area equation:
\[ \Delta = (-2ac \cos B) + 2ac \]
\[ \Delta = 2ac(1 - \cos B) \]
Recall the half-angle identity \( 1 - \cos B = 2 \sin^2 \frac{B}{2} \):
\[ \Delta = 2ac (2 \sin^2 \frac{B}{2}) = 4ac \sin^2 \frac{B}{2} \]
Now equate this to the standard area formula \( \frac{1}{2} ac \sin B \):
\[ \frac{1}{2} ac \sin B = 4ac \sin^2 \frac{B}{2} \]
Cancel \( ac \) from both sides (since they are side lengths and non-zero):
\[ \frac{1}{2} \sin B = 4 \sin^2 \frac{B}{2} \]
Use the identity \( \sin B = 2 \sin\frac{B}{2} \cos\frac{B}{2} \):
\[ \frac{1}{2} (2 \sin\frac{B}{2} \cos\frac{B}{2}) = 4 \sin^2 \frac{B}{2} \]
\[ \sin\frac{B}{2} \cos\frac{B}{2} = 4 \sin^2 \frac{B}{2} \]
Divide both sides by \( \sin\frac{B}{2} \cos\frac{B}{2} \):
\[ 1 = 4 \frac{\sin\frac{B}{2}}{\cos\frac{B}{2}} = 4 \tan \frac{B}{2} \]
\[ \tan \frac{B}{2} = \frac{1}{4} \]
Finally, calculate \( \tan B \) using the double-angle formula:
\[ \tan B = \frac{2 \tan\frac{B}{2}}{1 - \tan^2\frac{B}{2}} \]
\[ \tan B = \frac{2(1/4)}{1 - (1/4)^2} = \frac{1/2}{1 - 1/16} \]
\[ \tan B = \frac{1/2}{15/16} = \frac{1}{2} \times \frac{16}{15} = \frac{8}{15} \]
Step 4: Final Answer:
The value of \( \tan B \) is \( 8/15 \).
This matches Option (D).