The problem requires us to calculate the value of tan A for a triangle ABC, given a specific formula for its area, Δ. The provided area formula is Δ = a2 - (b - c)2.

Area Formula Analysis and Derivation

We will use standard triangle formulas to solve this. The area of a triangle can be expressed in terms of two sides and the included angle as Δ = \frac{1}{2} bc \sin A. Additionally, the Law of Cosines relates the sides and one angle: a2 = b2 + c2 - 2bc \cos A.

First, let's expand the given area formula:

Δ = a2 - (b - c)2

Expanding the squared term (b - c)2 gives:

Δ = a2 - (b2 - 2bc + c2)

Distributing the negative sign:

Δ = a2 - b2 - c2 + 2bc

Now, let's rearrange the Law of Cosines to find an expression for a2 - b2 - c2. From a2 = b2 + c2 - 2bc \cos A, we can write:

a2 - b2 - c2 = -2bc \cos A

Substitute this result back into our expanded area formula:

Δ = (-2bc \cos A) + 2bc

Factor out the common term 2bc:

Δ = 2bc (1 - \cos A)

Equating Area Expressions for tan A

We now have two different expressions for the area Δ:

• Δ = \frac{1}{2} bc \sin A (Standard formula)
• Δ = 2bc (1 - \cos A) (Derived formula)

By equating these two expressions, we get:

\frac{1}{2} bc \sin A = 2bc (1 - \cos A)

Assuming b and c are non-zero side lengths, we can cancel bc from both sides and multiply by 2:

\sin A = 4 (1 - \cos A)

This equation can be rewritten as:

\sin A = 4 - 4 \cos A

Or, rearranging:

\sin A + 4 \cos A = 4

Solving Using Half-Angle Formulas

To find tan A, we can utilize the tangent half-angle substitution. Let t = \tan(A/2). We use the identities:

• \sin A = \frac{2t}{1 + t^2}
• \cos A = \frac{1 - t^2}{1 + t^2}

Substitute these identities into the equation \sin A = 4(1 - \cos A):

\frac{2t}{1 + t^2} = 4 \left( 1 - \frac{1 - t^2}{1 + t^2} \right)

Simplify the right side:

\frac{2t}{1 + t^2} = 4 \left( \frac{(1 + t^2) - (1 - t^2)}{1 + t^2} \right)

\frac{2t}{1 + t^2} = 4 \left( \frac{1 + t^2 - 1 + t^2}{1 + t^2} \right)

\frac{2t}{1 + t^2} = 4 \left( \frac{2t^2}{1 + t^2} \right)

\frac{2t}{1 + t^2} = \frac{8t^2}{1 + t^2}

Since 1 + t^2 is never zero, we can multiply both sides by (1 + t^2):

2t = 8t^2

Rearrange this into a quadratic equation:

8t^2 - 2t = 0

Factor out 2t:

2t (4t - 1) = 0

This equation yields two possible solutions for t: t = 0 or t = 1/4.

If t = \tan(A/2) = 0, then A/2 would be 0 or \pi, implying A = 0 or A = 2\pi. These angles do not form a valid triangle. Therefore, we must use the other solution:

\tan(A/2) = \frac{1}{4}.

Calculating the Final tan A Value

Now, we use the double angle formula for tangent, which relates tan A to tan(A/2):

\tan A = \frac{2 \tan(A/2)}{1 - \tan^2(A/2)}

Substitute the value \tan(A/2) = \frac{1}{4} into the formula:

\tan A = \frac{2 \times \frac{1}{4}}{1 - (\frac{1}{4})^2}

\tan A = \frac{\frac{1}{2}}{1 - \frac{1}{16}}

Simplify the denominator:

\tan A = \frac{\frac{1}{2}}{\frac{16 - 1}{16}}

\tan A = \frac{\frac{1}{2}}{\frac{15}{16}}

To divide by a fraction, we multiply by its reciprocal:

\tan A = \frac{1}{2} \times \frac{16}{15}

\tan A = \frac{16}{30}

Simplify the resulting fraction by dividing the numerator and denominator by 2:

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

Final tan A Result

The calculated value of tan A is \frac{8}{15}.