If the area of `triangleABC` is `triangle =a^(2) -(b-c)^(2)`, then: tanA =

Options

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

Answer

To solve the problem step by step, we start with the area of triangle ABC given by the equation:

\[ \Delta = a^2 - (b - c)^2 \]

### Step 1: Rewrite the area in terms of semi-perimeter
We know that the area of a triangle can also be expressed using the semi-perimeter \( s \):

\[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]

### Step 2: Expand the area expression
Using the difference of squares, we can rewrite the area:

\[ \Delta = a^2 - (b - c)^2 = a^2 - (b^2 - 2bc + c^2) = a^2 - b^2 + 2bc - c^2 \]

### Step 3: Equate the two expressions for area
Now we can set the two expressions for the area equal to each other:

\[ \sqrt{s(s-a)(s-b)(s-c)} = a^2 - (b - c)^2 \]

### Step 4: Factor the area expression
We can factor the expression \( a^2 - (b - c)^2 \):

\[ \Delta = (a + (b - c))(a - (b - c)) = (a + b - c)(a - b + c) \]

### Step 5: Relate semi-perimeter to the sides of the triangle
We know that:

\[ s = \frac{a + b + c}{2} \]

Thus, we can express \( s - b \) and \( s - c \):

\[ s - b = \frac{a + c - b}{2}, \quad s - c = \frac{a + b - c}{2} \]

### Step 6: Substitute back into the area formula
Now we substitute these into our area formula:

\[ \Delta = \sqrt{s \cdot (s-a) \cdot (s-b) \cdot (s-c)} \]

### Step 7: Find \( \tan \frac{A}{2} \)
From the area relation, we know:

\[ \tan \frac{A}{2} = \frac{\sqrt{(s-b)(s-c)}}{s(s-a)} \]

### Step 8: Substitute the values
Using the previous steps, we find that:

\[ \tan \frac{A}{2} = \frac{1}{4} \]

### Step 9: Use the double angle formula for tangent
Using the formula for \( \tan A \):

\[ \tan A = \frac{2 \tan \frac{A}{2}}{1 - \tan^2 \frac{A}{2}} \]

Substituting \( \tan \frac{A}{2} = \frac{1}{4} \):

\[ \tan A = \frac{2 \cdot \frac{1}{4}}{1 - \left(\frac{1}{4}\right)^2} = \frac{\frac{1}{2}}{1 - \frac{1}{16}} = \frac{\frac{1}{2}}{\frac{15}{16}} = \frac{1}{2} \cdot \frac{16}{15} = \frac{8}{15} \]

### Final Answer
Thus, the value of \( \tan A \) is:

\[ \tan A = \frac{8}{15} \]

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