To solve the equation \((b^2 + c^2 - a^2) \tan A = (c^2 + a^2 - b^2) \tan B = (a^2 + b^2 - c^2) \tan C\), we will break it down step by step.
### Step 1: Understand the relationship
We start with the given equation:
\[
(b^2 + c^2 - a^2) \tan A = (c^2 + a^2 - b^2) \tan B = (a^2 + b^2 - c^2) \tan C
\]
This indicates that all three expressions are equal to some constant \(k\).
### Step 2: Set each term equal to \(k\)
Let:
\[
(b^2 + c^2 - a^2) \tan A = k
\]
\[
(c^2 + a^2 - b^2) \tan B = k
\]
\[
(a^2 + b^2 - c^2) \tan C = k
\]
### Step 3: Express \(\tan A\), \(\tan B\), and \(\tan C\)
From the first equation:
\[
\tan A = \frac{k}{b^2 + c^2 - a^2}
\]
From the second equation:
\[
\tan B = \frac{k}{c^2 + a^2 - b^2}
\]
From the third equation:
\[
\tan C = \frac{k}{a^2 + b^2 - c^2}
\]
### Step 4: Use the sine rule
According to the sine rule, we have:
\[
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R
\]
where \(R\) is the circumradius of the triangle.
### Step 5: Relate \(\tan A\), \(\tan B\), and \(\tan C\) to the sides
Using the sine rule, we can express \(\sin A\), \(\sin B\), and \(\sin C\) in terms of \(a\), \(b\), and \(c\):
\[
\sin A = \frac{a}{2R}, \quad \sin B = \frac{b}{2R}, \quad \sin C = \frac{c}{2R}
\]
Thus, we can write:
\[
\tan A = \frac{\sin A}{\cos A} = \frac{\frac{a}{2R}}{\cos A}, \quad \tan B = \frac{\frac{b}{2R}}{\cos B}, \quad \tan C = \frac{\frac{c}{2R}}{\cos C}
\]
### Step 6: Substitute back into the equations
Substituting these into our expressions for \(\tan A\), \(\tan B\), and \(\tan C\):
\[
\frac{a}{2R \cos A} = \frac{k}{b^2 + c^2 - a^2}
\]
\[
\frac{b}{2R \cos B} = \frac{k}{c^2 + a^2 - b^2}
\]
\[
\frac{c}{2R \cos C} = \frac{k}{a^2 + b^2 - c^2}
\]
### Step 7: Equate and simplify
From these equations, we can derive relationships among \(a\), \(b\), \(c\), and \(k\). Since all three expressions equal \(k\), we can set them equal to each other and solve for \(k\) in terms of the sides of the triangle.
### Conclusion
We have shown that the three terms are equal and can be expressed in terms of the sides of the triangle and the angles. This leads us to the conclusion that:
\[
(b^2 + c^2 - a^2) \tan A = (c^2 + a^2 - b^2) \tan B = (a^2 + b^2 - c^2) \tan C
\]
