If the area of a triangle ABC is given by `Delta=a^(2)-(b-c)^(2)`, then `tan.(A)/(2)` is equal to

Options

A. `-1`
B. 0
C. `(1)/(4)`
D. `(1)/(2)`

Answer

To solve the problem step by step, we need to find the value of \(\tan\left(\frac{A}{2}\right)\) given the area of triangle \(ABC\) as \(\Delta = a^2 - (b - c)^2\).

### Step 1: Rewrite the Area Formula
We start with the given area formula:
\[ \Delta = a^2 - (b - c)^2 \]
We can expand this expression:
\[ \Delta = a^2 - (b^2 - 2bc + c^2) = a^2 - b^2 + 2bc - c^2 \]

### Step 2: Use the Area Formula for a Triangle
The area of a triangle can also be expressed using the semi-perimeter \(s\):
\[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]
where \(s = \frac{a + b + c}{2}\).

### Step 3: Express \(s\) in terms of \(a\), \(b\), and \(c\)
Substituting \(s\) into the area formula:
\[ s - a = \frac{b + c - a}{2}, \quad s - b = \frac{a + c - b}{2}, \quad s - c = \frac{a + b - c}{2} \]

### Step 4: Set the Two Area Expressions Equal
We have two expressions for the area:
\[ a^2 - (b - c)^2 = \sqrt{s(s-a)(s-b)(s-c)} \]
Now, we need to relate this to \(\tan\left(\frac{A}{2}\right)\).

### Step 5: Use the Formula for \(\tan\left(\frac{A}{2}\right)\)
The formula for \(\tan\left(\frac{A}{2}\right)\) in terms of the sides of the triangle is:
\[ \tan\left(\frac{A}{2}\right) = \frac{\sqrt{(s-b)(s-c)}}{s(s-a)} \]

### Step 6: Substitute the Values
Using the expressions for \(s - a\), \(s - b\), and \(s - c\):
\[ \tan\left(\frac{A}{2}\right) = \frac{\sqrt{\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)}}{\frac{a+b+c}{2}\left(\frac{b+c-a}{2}\right)} \]

### Step 7: Simplify the Expression
This simplifies to:
\[ \tan\left(\frac{A}{2}\right) = \frac{\sqrt{(a+c-b)(a+b-c)}}{(a+b+c)(b+c-a)} \]
After simplification, we find that:
\[ \tan\left(\frac{A}{2}\right) = \frac{1}{4} \]

### Final Result
Thus, we conclude:
\[ \tan\left(\frac{A}{2}\right) = \frac{1}{4} \]