For any triangle ABC, prove that`(b^2 - c^2) cotA + (c^2 - a^2) cotB + (a^2 - b^2) cotC = 0`

Explanation

To prove the identity \((b^2 - c^2) \cot A + (c^2 - a^2) \cot B + (a^2 - b^2) \cot C = 0\) for any triangle \(ABC\), we will use the sine and cosine rules.

### Step 1: Use the Sine Rule
The sine rule states that:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \quad \text{(some constant)} \]
From this, we can express \(\sin A\), \(\sin B\), and \(\sin C\) in terms of \(a\), \(b\), and \(c\):
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