If in a triangle `ABC, S = a^2 - (b -c)^2` then tan A is equal to

Options

A. `1/2`
B. `8/17`
C. `8/15`
D. `15/16`

Answer

To solve the problem, we need to find the value of \(\tan A\) in triangle \(ABC\) given that \(S = a^2 - (b - c)^2\).

### Step-by-Step Solution:

1. **Understanding the Given Expression**:
We start with the expression given for \(S\):
\[ S = a^2 - (b - c)^2 \]
This can be simplified using the identity \(x^2 - y^2 = (x - y)(x + y)\):
\[ S = a^2 - (b^2 - 2bc + c^2) = a^2 + 2bc - b^2 - c^2 \]

2. **Using the Area of Triangle**:
The area \(S\) of triangle \(ABC\) can also be expressed using Heron's formula:
\[ S = \sqrt{s(s-a)(s-b)(s-c)} \]
where \(s = \frac{a+b+c}{2}\).

3. **Setting Up the Equation**:
Equating the two expressions for \(S\):
\[ \sqrt{s(s-a)(s-b)(s-c)} = a^2 + 2bc - b^2 - c^2 \]

4. **Finding \(\tan A\)**:
We know that:
\[ \tan A = \frac{a}{h} \]
where \(h\) is the height from vertex \(A\) to side \(BC\). The area can also be expressed as:
\[ S = \frac{1}{2} \times b \times c \times \sin A \]
Thus, we can relate the area to \(\tan A\):
\[ S = \frac{1}{2} a h \]
From this, we can express \(h\) in terms of \(S\):
\[ h = \frac{2S}{a} \]
Therefore, substituting this into the expression for \(\tan A\):
\[ \tan A = \frac{a}{\frac{2S}{a}} = \frac{a^2}{2S} \]

5. **Substituting for \(S\)**:
Now substitute \(S = a^2 - (b - c)^2\) into the expression for \(\tan A\):
\[ \tan A = \frac{a^2}{2(a^2 - (b - c)^2)} \]

### Final Expression:
Thus, the final expression for \(\tan A\) is:
\[ \tan A = \frac{a^2}{2(a^2 - (b - c)^2)} \]