<Question>
The area of triangle $ABC$ is equal to $a^2 - (b - c)^2,$ where $a,$ $b,$ and $c$ are the sides of triangle $ABC,$ as usual.  Compute $\tan A.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the tangent of angle A in triangle ABC, given that the area of the triangle is equal to \( a^2 - (b - c)^2 \). Hmm, let me start by recalling some formulas related to the area of a triangle and trigonometric identities.

First, the area of a triangle can be expressed in several ways. The most common one is \( \frac{1}{2}ab\sin C \), where a and b are two sides and C is the included angle. In this case, since the sides are labeled a, b, and c, and we need to find tan A, which is related to angle A, I should probably use the formula involving sides b and c and angle A. So the area would be \( \frac{1}{2}bc\sin A \).

<prompt_1.1> Express the area of triangle ABC using the formula involving sides b, c, and angle A. </prompt_1.1>

But let's verify if this is possible. Let's check the discriminant: \( 16c^2 - 4 \times 1 \times c^2 = 16c^2 - 4c^2 = 12c^2 \), which is positive, so there are two real roots.

Wait, but this seems odd. How can angle A be determined uniquely from this equation? Maybe there's another approach.
-----

Alternatively, maybe there's a mistake in the problem statement? Wait, no, the problem states the area is equal to \( a^2 - (b - c)^2 \). Let me check the problem again.
-----

So perhaps the answer is 2?

But let me check with another method to confirm. Let's use the Law of Cosines. The Law of Cosines states that \( a^2 = b^2 + c^2 - 2bc\cos A \).
Rearranging, we get \( \cos A = \frac{b^2 + c^2 - a^2}{2bc} \). Then, \( \tan A = \frac{\sin A}{\cos A} = \frac{\frac{2\Delta}{bc}}{\frac{b^2 + c^2 - a^2}{2bc}} = \frac{4\Delta}{b^2 + c^2 - a^2} \).

<prompt_1.2> Use the Law of Cosines and the expression for \(\tan A\) to find the final answer. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
Wait, the formula is \( \frac{1}{2} \times \text{base} \times \text{height} \), but in terms of two sides and the included angle, yes, it's \( \frac{1}{2}bc\sin A \). So the area is \( \frac{1}{2}bc\sin A \).
</execution_1.1>

<execution_1.2>
Given that \( \Delta = a^2 - (b - c)^2 \), substitute this into the equation:

\( \tan A = \frac{4(a^2 - (b - c)^2)}{b^2 + c^2 - a^2} \)

Let me simplify the numerator:

\( 4(a^2 - (b^2 - 2bc + c^2)) = 4(a^2 - b^2 + 2bc - c^2) = 4a^2 - 4b^2 + 8bc - 4c^2 \)

The denominator is \( b^2 + c^2 - a^2 \), so let's see if we can relate the numerator and denominator.

Wait, perhaps there's a different approach. Let's use the area formula again. We have \( \Delta = a^2 - (b - c)^2 \), which is also equal to \( \frac{1}{2}bc\sin A \). So equate them:

\( \frac{1}{2}bc\sin A = a^2 - (b - c)^2 \)

Then, \( \sin A = \frac{2(a^2 - (b - c)^2)}{bc} \). Then, \( \tan A = \frac{\sin A}{\cos A} \). But we need to find \( \cos A \) as well.

From the Law of Cosines, \( \cos A = \frac{b^2 + c^2 - a^2}{2bc} \). So then:

\( \tan A = \frac{\frac{2(a^2 - (b - c)^2)}{bc}}{\frac{b^2 + c^2 - a^2}{2bc}} = \frac{4(a^2 - (b - c)^2)}{b^2 + c^2 - a^2} \)

But this seems complicated. Maybe there's a way to simplify this expression. Let's expand the numerator:

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

So the numerator becomes:

\( 4(a^2 - b^2 + 2bc - c^2) = 4a^2 - 4b^2 + 8bc - 4c^2 \)

The denominator is \( b^2 + c^2 - a^2 \). Let's see if we can factor or relate these terms.

Alternatively, perhaps there's a mistake in my approach. Let me think again.

Wait, maybe instead of using the Law of Cosines, I can use another identity. Let's recall that \( \tan A = \frac{\sin A}{\cos A} \). We already have \( \sin A = \frac{2\Delta}{bc} \), so if I can express \( \cos A \) in terms of the sides, then I can find \( \tan A \).

Alternatively, perhaps there's a way to express \( \tan A \) directly in terms of the sides. Let me think.

Wait, perhaps using Heron's formula? But the area is given as \( a^2 - (b - c)^2 \), which might not be expressible directly via Heron's formula. Let me check.

Heron's formula states that the area is \( \sqrt{s(s - a)(s - b)(s - c)} \), where \( s = \frac{a + b + c}{2} \). But given that the area here is \( a^2 - (b - c)^2 \), unless this expression can be related to Heron's formula, which seems unlikely, perhaps this approach isn't helpful.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me set them equal:

\( \frac{1}{2}bc\sin A = a^2 - (b - c)^2 \)

I need to solve for \( \sin A \) here.

<prompt_2.1> Solve the equation \( \frac{1}{2}bc\sin A = a^2 - (b - c)^2 \) for \( \sin A \). </prompt_2.1>
-----

Based on execution_1.2:
Let me try plugging in some numbers to see if I can get a pattern.

Suppose we take a specific case where the triangle is right-angled at A.

<prompt_2.2> Test the expression for \(\tan A\) by considering a right-angled triangle at A and checking if \(\tan A = 2\). </prompt_2.2>
-----

Based on execution_1.1:
Wait, but how does that help me find tan A? Let me think.

I need another equation that relates the sides a, b, c. Maybe use the Law of Cosines? Since we have angle A here, the Law of Cosines states that \( a^2 = b^2 + c^2 - 2bc\cos A \).

<prompt_2.3> Use the Law of Cosines to find an expression for \(\tan A\) in terms of the sides of the triangle. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
First, expand the right-hand side:

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

So, the equation becomes:

\( \frac{1}{2}bc\sin A = a^2 - b^2 + 2bc - c^2 \)

Multiply both sides by 2 to eliminate the fraction:

\( bc\sin A = 2a^2 - 2b^2 + 4bc - 2c^2 \)

Then, divide both sides by \( bc \):

\( \sin A = \frac{2a^2 - 2b^2 + 4bc - 2c^2}{bc} \)

Hmm, that seems a bit complicated. Maybe there's a way to simplify this expression. Let me factor the numerator:

\( 2a^2 - 2b^2 + 4bc - 2c^2 = 2(a^2 - b^2 + 2bc - c^2) \)

So,

\( \sin A = \frac{2(a^2 - b^2 + 2bc - c^2)}{bc} \)
</execution_2.1>

<execution_2.2>
Then, by Pythagoras, \( a^2 = b^2 + c^2 \). Let's see if this satisfies the given area formula.

Compute \( a^2 - (b - c)^2 = (b^2 + c^2) - (b^2 - 2bc + c^2) = b^2 + c^2 - b^2 + 2bc - c^2 = 2bc \).

The area of a right-angled triangle at A is \( \frac{1}{2}bc \). So according to the problem's area formula, it should be \( 2bc \). But in reality, the area is \( \frac{1}{2}bc \). This suggests a contradiction unless \( \frac{1}{2}bc = 2bc \), which would imply \( \frac{1}{2} = 2 \), which is false. Therefore, angle A is not right. So that case is impossible.
</execution_2.2>

<execution_2.3>
So if I can express \( \cos A \) in terms of the sides, and I already have an expression for \( \sin A \), then \( \tan A = \frac{\sin A}{\cos A} \).

Given that the area is \( \frac{1}{2}bc\sin A = a^2 - (b - c)^2 \), so \( \sin A = \frac{2(a^2 - (b - c)^2)}{bc} \). Then, from the Law of Cosines, \( \cos A = \frac{b^2 + c^2 - a^2}{2bc} \).

Therefore, \( \tan A = \frac{\frac{2(a^2 - (b - c)^2)}{bc}}{\frac{b^2 + c^2 - a^2}{2bc}} = \frac{4(a^2 - (b - c)^2)}{b^2 + c^2 - a^2} \).
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.3:
Let me simplify the numerator:

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

So, the numerator becomes \( 4(a^2 - b^2 + 2bc - c^2) \), and the denominator is \( b^2 + c^2 - a^2 \).

So, \( \tan A = \frac{4(a^2 - b^2 + 2bc - c^2)}{b^2 + c^2 - a^2} \)

Hmm, perhaps we can factor the numerator and denominator.

<prompt_3.1> Simplify the expression for \(\tan A\) by factoring the numerator and denominator. </prompt_3.1>
-----

Based on execution_2.1:
Let's recall that \( \sin^2 A + \cos^2 A = 1 \). Maybe if I can find \( \cos A \), then I can find \( \tan A \).
Alternatively, perhaps relate \( \sin A \) and \( \cos A \) through some identity. Wait, but I need another equation involving \( \sin A \) and \( \cos A \). Let's think.

<prompt_3.2> Use the Pythagorean identity to express \(\tan A\) in terms of \(\sin A\) and \(\cos A\). </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Let me rearrange the terms in the numerator:

\( 4(-b^2 - c^2 + a^2 + 2bc + 2bc) \). Wait, no, let's see:

Wait, the numerator is \( 4(a^2 - b^2 + 2bc - c^2) = 4[a^2 - (b^2 + c^2) + 2bc] \). The denominator is \( b^2 + c^2 - a^2 \). Let me denote \( D = b^2 + c^2 - a^2 \). Then the numerator is \( 4[-D + 2bc] = 4(2bc - D) \). Therefore, \( \tan A = \frac{4(2bc - D)}{D} = \frac{8bc - 4D}{D} = \frac{8bc}{D} - 4 \). But this might not be helpful.

Alternatively, let's write the numerator as \( 4(a^2 - b^2 - c^2 + 2bc) = 4[(a^2) - (b^2 + c^2 - 2bc)] = 4[a^2 - (b - c)^2] \). Wait, but that's just \( 4 \times \) the original area expression. Since the area is \( a^2 - (b - c)^2 \), which is equal to \( \frac{1}{2}bc\sin A \). Therefore, the numerator is \( 4 \times \frac{1}{2}bc\sin A = 2bc\sin A \). Therefore, \( \tan A = \frac{2bc\sin A}{b^2 + c^2 - a^2} \).

But from the area, we have \( \frac{1}{2}bc\sin A = a^2 - (b - c)^2 \), so \( bc\sin A = 2(a^2 - (b - c)^2) \). Substitute this into the expression for \( \tan A \):

\( \tan A = \frac{2 \times 2(a^2 - (b - c)^2)}{b^2 + c^2 - a^2} = \frac{4(a^2 - (b - c)^2)}{b^2 + c^2 - a^2} \)

Which is the same as before.
</execution_3.1>

<execution_3.2>
From the Law of Cosines, we have \( \cos A = \frac{b^2 + c^2 - a^2}{2bc} \). So, if I can express \( \sin A \) in terms of \( \cos A \), then maybe I can find \( \tan A \). Let's see.

We know that \( \sin A = \sqrt{1 - \cos^2 A} \), but that might complicate things. Alternatively, since we have an expression for \( \sin A \) from earlier, maybe we can use that to find \( \tan A \).

Wait, but \( \tan A = \frac{\sin A}{\cos A} \), so if we have expressions for both \( \sin A \) and \( \cos A \), we can just divide them. Let's try that.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
So, \( \sin A = \frac{2(a^2 - (b - c)^2)}{bc} \) from earlier. And \( \cos A = \frac{b^2 + c^2 - a^2}{2bc} \).

<prompt_4.1> Divide the expressions for \(\sin A\) and \(\cos A\) to find \(\tan A\). </prompt_4.1>
-----

Based on execution_3.1, execution_3.2:
But how can angle A have a tangent of 2? That would mean angle A is arctan(2), which is approximately 63.43 degrees.
Since tangent is positive, angle A must be acute.

Wait, but in a triangle, angles can be acute or obtuse, but if tangent is 2, it's acute. Let's check if this makes sense.

<prompt_4.2> Verify if the result for \(\tan A = 2\) is consistent with the properties of a triangle. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
Therefore, \( \tan A = \frac{\frac{2(a^2 - (b - c)^2)}{bc}}{\frac{b^2 + c^2 - a^2}{2bc}} = \frac{4(a^2 - (b - c)^2)}{b^2 + c^2 - a^2} \)
</execution_4.1>

<execution_4.2>
Let me verify with the earlier assumption where angle A was right, but that led to a contradiction. So if angle A is arctan(2), that's a valid acute angle.

Alternatively, perhaps there's a way to confirm this answer. Let me think of another approach.

Suppose we use Heron's formula. The area is given as \( a^2 - (b - c)^2 \). Let's compute the semi-perimeter:

\( s = \frac{a + b + c}{2} \)

Then, Heron's formula says area is \( \sqrt{s(s - a)(s - b)(s - c)} \). But this seems complicated to equate to \( a^2 - (b - c)^2 \). Maybe instead, consider expanding \( a^2 - (b - c)^2 \):

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

So, the area is \( a^2 - b^2 - c^2 + 2bc \). Hmm, but how does this relate to the standard area formulas? Alternatively, perhaps use the formula \( \Delta = \frac{1}{2}ab\sin C \), but we already did that.

Alternatively, maybe express \( a^2 \) in terms of \( b \) and \( c \) using the Law of Cosines. From the Law of Cosines:

\( a^2 = b^2 + c^2 - 2bc\cos A \)

Substitute this into the area expression:

\( \Delta = (b^2 + c^2 - 2bc\cos A) - (b - c)^2 \)

Expand \( (b - c)^2 = b^2 - 2bc + c^2 \), so:

\( \Delta = b^2 + c^2 - 2bc\cos A - b^2 + 2bc - c^2 = -2bc\cos A + 2bc = 2bc(1 - \cos A) \)

But we also know that \( \Delta = \frac{1}{2}bc\sin A \). Therefore, equate the two expressions for the area:

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

Divide both sides by \( bc \):

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

Multiply both sides by 2:

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

Now, recall the identity \( \sin A = 2\sin\frac{A}{2}\cos\frac{A}{2} \) and \( 1 - \cos A = 2\sin^2\frac{A}{2} \). Substitute these into the equation:

\( 2\sin\frac{A}{2}\cos\frac{A}{2} = 4 \times 2\sin^2\frac{A}{2} \)

Simplify the right-hand side:

\( 2\sin\frac{A}{2}\cos\frac{A}{2} = 8\sin^2\frac{A}{2} \)

Divide both sides by \( 2\sin\frac{A}{2} \) (assuming \( \sin\frac{A}{2} \neq 0 \), which is true since angle A is between 0 and 180 degrees, so \( \frac{A}{2} \) is between 0 and 90 degrees):

\( \cos\frac{A}{2} = 4\sin\frac{A}{2} \)

Divide both sides by \( \sin\frac{A}{2} \):

\( \cot\frac{A}{2} = 4 \)

Therefore,

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

Now, we can use the double-angle formula for tangent to find \( \tan A \). Recall that:

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

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

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

Wait a minute, this contradicts the previous result where I thought \( \tan A = 2 \). Hmm, so which one is correct?

Let me check my steps again.

Starting from \( \sin A = 4(1 - \cos A) \). Then, using the half-angle identities:

\( \sin A = 2\sin\frac{A}{2}\cos\frac{A}{2} \)

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

So, substituting:

\( 2\sin\frac{A}{2}\cos\frac{A}{2} = 8\sin^2\frac{A}{2} \)

Divide both sides by \( 2\sin\frac{A}{2} \):

\( \cos\frac{A}{2} = 4\sin\frac{A}{2} \)

Which leads to \( \cot\frac{A}{2} = 4 \), so \( \tan\frac{A}{2} = \frac{1}{4} \). Then, using the double-angle formula:

\( \tan A = \frac{2 \times \frac{1}{4}}{1 - (\frac{1}{4})^2} = \frac{\frac{1}{2}}{\frac{15}{16}} = \frac{8}{15} \)

So, \( \tan A = \frac{8}{15} \).
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.2:
Wait, but this contradicts the earlier result where I thought \( \tan A = 2 \). So there must be a mistake in one of the approaches. Let's check both methods again.

<prompt_5.1> Compare the results from both methods to identify any contradictions. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
First method:

1. Start with area \( \Delta = \frac{1}{2}bc\sin A = a^2 - (b - c)^2 \)
2. Express \( \sin A \) in terms of \( a, b, c \)
3. Use Law of Cosines to express \( \cos A \) in terms of sides
4. Compute \( \tan A = \frac{\sin A}{\cos A} \), leading to \( \tan A = 2 \)

Second method:

1. Express the given area \( a^2 - (b - c)^2 \) as \( a^2 - b^2 - c^2 + 2bc \)
2. Equate to Heron's formula and use Law of Cosines
3. Derive equation \( \sin A = 4(1 - \cos A) \)
4. Solve using half-angle identities, leading to \( \tan A = \frac{8}{15} \)

There's a discrepancy here. Let me check the second method again.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Starting from the area expression:

<prompt_6.1> Verify the second method from the start to identify any errors. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
\( \Delta = a^2 - (b - c)^2 \)

Expand \( (b - c)^2 = b^2 - 2bc + c^2 \), so:

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

But also, from Law of Cosines:

\( a^2 = b^2 + c^2 - 2bc\cos A \)

Therefore, substituting into \( \Delta \):

\( \Delta = (b^2 + c^2 - 2bc\cos A) - (b^2 - 2bc + c^2) \)

Simplify:

\( \Delta = b^2 + c^2 - 2bc\cos A - b^2 + 2bc - c^2 = -2bc\cos A + 2bc = 2bc(1 - \cos A) \)

But also, \( \Delta = \frac{1}{2}bc\sin A \). Therefore:

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

Divide both sides by \( bc \):

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

Multiply both sides by 2:

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

Yes, that's correct. Then, using the identity \( \sin A = 2\sin\frac{A}{2}\cos\frac{A}{2} \) and \( 1 - \cos A = 2\sin^2\frac{A}{2} \), substituting gives:

\( 2\sin\frac{A}{2}\cos\frac{A}{2} = 8\sin^2\frac{A}{2} \)

Divide by \( 2\sin\frac{A}{2} \):

\( \cos\frac{A}{2} = 4\sin\frac{A}{2} \implies \cot\frac{A}{2} = 4 \implies \tan\frac{A}{2} = \frac{1}{4} \)

Then, using the double-angle formula for tangent:

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

Therefore, \( \tan A = \frac{8}{15} \). But in the first method, I arrived at \( \tan A = 2 \). Clearly, there is a contradiction here. Which one is correct?

Wait, let's check the first method again.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
First method:

Starting with \( \Delta = a^2 - (b - c)^2 \)

Express \( a^2 \) as \( b^2 + c^2 - 2bc\cos A \), so:

\( \Delta = (b^2 + c^2 - 2bc\cos A) - (b^2 - 2bc + c^2) = -2bc\cos A + 2bc = 2bc(1 - \cos A) \)

But also, \( \Delta = \frac{1}{2}bc\sin A \), so equate:

\( \frac{1}{2}bc\sin A = 2bc(1 - \cos A) \implies \sin A = 4(1 - \cos A) \)

Then, using \( \sin A = 4(1 - \cos A) \), which leads to \( \tan A = \frac{8}{15} \). Wait, but how?

Wait, perhaps I made a mistake in the first method when I tried to compute \( \tan A \). Let me re-examine that step.

<prompt_7.1> Re-evaluate the first method to identify any errors and confirm the correct result. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
If \( \sin A = 4(1 - \cos A) \), then squaring both sides:

\( \sin^2 A = 16(1 - 2\cos A + \cos^2 A) \)

But \( \sin^2 A = 1 - \cos^2 A \), so:

\( 1 - \cos^2 A = 16 - 32\cos A + 16\cos^2 A \)

Bring all terms to one side:

\( 1 - \cos^2 A - 16 + 32\cos A - 16\cos^2 A = 0 \)

Combine like terms:

\( -17\cos^2 A + 32\cos A - 15 = 0 \)

Multiply by -1:

\( 17\cos^2 A - 32\cos A + 15 = 0 \)

Let \( x = \cos A \), then:

\( 17x^2 - 32x + 15 = 0 \)

Solve using quadratic formula:

\( x = \frac{32 \pm \sqrt{(-32)^2 - 4 \times 17 \times 15}}{2 \times 17} = \frac{32 \pm \sqrt{1024 - 1020}}{34} = \frac{32 \pm \sqrt{4}}{34} = \frac{32 \pm 2}{34} \)

Thus, \( x = \frac{34}{34} = 1 \) or \( x = \frac{30}{34} = \frac{15}{17} \)

\( \cos A = 1 \) implies \( A = 0^\circ \), which is impossible for a triangle. Therefore, \( \cos A = \frac{15}{17} \), so \( \sin A = \sqrt{1 - \left(\frac{15}{17}\right)^2} = \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{64}{289}} = \frac{8}{17} \)

Therefore, \( \tan A = \frac{\sin A}{\cos A} = \frac{\frac{8}{17}}{\frac{15}{17}} = \frac{8}{15} \)

So, this confirms the result from the second method. Therefore, \( \tan A = \frac{8}{15} \). But in the first method, when I tried to compute \( \tan A \) directly from \( \sin A = 4(1 - \cos A) \), I must have made a mistake.

Wait, in the first method, when I tried to compute \( \tan A = \frac{\sin A}{\cos A} \), I substituted \( \sin A = 4(1 - \cos A) \), so \( \tan A = \frac{4(1 - \cos A)}{\cos A} = 4\left(\frac{1}{\cos A} - 1\right) \). Then, using \( \cos A = \frac{15}{17} \), substitute:

\( \tan A = 4\left(\frac{17}{15} - 1\right) = 4\left(\frac{2}{15}\right) = \frac{8}{15} \), which matches the second method. So, in the first method, after finding \( \cos A = \frac{15}{17} \), we can directly compute \( \tan A \). Therefore, both methods agree that \( \tan A = \frac{8}{15} \).

But earlier, when I tried to express \( \tan A \) in terms of \( \sin A \) and \( \cos A \) without substituting the value of \( \cos A \), I might have confused the expressions. However, by solving the quadratic equation, we find the exact value of \( \cos A \), which then allows us to compute \( \tan A \) accurately.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
Therefore, the value of \( \tan A \) is \( \boxed{\dfrac{8}{15}} \).
**Final Answer**
\boxed{\dfrac{8}{15}}
</Plan_8>

<Execution_8>
No prompt to execute.
</Execution_8>

<Final_answer>
Therefore, the value of $\tan A$ is $\boxed{\frac{8}{15}}$.
</Final_answer>
</think>