Step 1: Understanding the Concept:
The problem provides an algebraic expression for the area (\( \Delta \)) of a triangle.
We must relate this algebraic form to the standard trigonometric formula for the area: \( \Delta = \frac{1}{2} ac \sin B \).
Additionally, the Cosine Rule relates the sides \( a, b, c \) to the cosine of the angle \( B \).
Key Formula or Approach:
1. Area of triangle: \( \Delta = \frac{1}{2} ac \sin B \).
2. Law of Cosines: \( b^2 = a^2 + c^2 - 2ac \cos B \).
3. Algebraic expansion: \( (c-a)^2 = c^2 + a^2 - 2ac \).
Step 2: Detailed Explanation:
The given area is \( \Delta = b^2 - (c - a)^2 \).
Expanding the square term:
\[ \Delta = b^2 - (c^2 + a^2 - 2ac) = b^2 - c^2 - a^2 + 2ac \]
Substitute the Law of Cosines expression \( b^2 = a^2 + c^2 - 2ac \cos B \):
\[ \Delta = (a^2 + c^2 - 2ac \cos B) - c^2 - a^2 + 2ac \]
The terms \( a^2 \) and \( c^2 \) cancel out:
\[ \Delta = 2ac - 2ac \cos B = 2ac (1 - \cos B) \]
We know that the area is also given by \( \Delta = \frac{1}{2} ac \sin B \).
Equating the two expressions for \( \Delta \):
\[ \frac{1}{2} ac \sin B = 2ac (1 - \cos B) \]
Assuming \( a, c \neq 0 \), we divide both sides by \( ac \):
\[ \frac{1}{2} \sin B = 2(1 - \cos B) \implies \sin B = 4(1 - \cos B) \]
Using half-angle identities \( \sin B = 2 \sin\frac{B}{2} \cos\frac{B}{2} \) and \( 1 - \cos B = 2 \sin^2\frac{B}{2} \):
\[ 2 \sin\frac{B}{2} \cos\frac{B}{2} = 4 (2 \sin^2\frac{B}{2}) = 8 \sin^2\frac{B}{2} \]
Divide by \( 2 \sin\frac{B}{2} \) (since \( B>0 \), \( \sin\frac{B}{2} \neq 0 \)):
\[ \cos\frac{B}{2} = 4 \sin\frac{B}{2} \implies \tan\frac{B}{2} = \frac{1}{4} \]
To find \( \tan B \), use the double angle formula:
\[ \tan B = \frac{2 \tan(B/2)}{1 - \tan^2(B/2)} = \frac{2(1/4)}{1 - (1/4)^2} = \frac{1/2}{1 - 1/16} \]
\[ \tan B = \frac{1/2}{15/16} = \frac{1}{2} \cdot \frac{16}{15} = \frac{8}{15} \]
Step 3: Final Answer:
The value of \( \tan B \) is \( 8/15 \).
This is Option (D).