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Question
In triangle ABC, prove the following:
Solution
Let
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]
Then,
Consider the LHS of the equation
\[LHS = \frac{\cos^2 B - \cos^2 C}{b + c} + \frac{\cos^2 C - \cos^2 A}{c + a} + \frac{\cos^2 A - \cos^2 B}{a + b}\]
\[Now, \]
\[\frac{\cos^2 B - \cos^2 C}{b + c} = \frac{\cos^2 B - \cos^2 C}{k\left( \sin B + \sin C \right)}\]
\[ = \frac{\left( \cos B + \cos C \right)\left( \cos B - cos C \right)}{k\left( \sin B + \sin C \right)} \left( \because \cos^2 B - \cos^2 C = \left( \cos B + \cos C \right)\left( \cos B - \cos C \right) \right)\]
\[ = \frac{\left[ 2\cos\left( \frac{B + C}{2} \right)\cos\left( \frac{B - C}{2} \right) \right]\left[ - 2\sin\left( \frac{B + C}{2} \right)\sin\left( \frac{B - C}{2} \right) \right]}{2k\sin\left( \frac{B + C}{2} \right)\sin\left( \frac{B - C}{2} \right)} \]
\[ = \frac{- 2\cos\left( \frac{B + C}{2} \right)\sin\left( \frac{B - C}{2} \right)}{k} = \frac{- \left( \sin B - \sin C \right)}{k} = \frac{\sin C - \sin B}{k}\]
Also,
\[\frac{\cos^2 C - \cos^2 A}{c + a} = \frac{\cos^2 C - \cos^2 A}{k\left( \sin C + \sin A \right)}\]
\[ = \frac{\left( \cos C + \cos A \right)\left( \cos C - \cos A \right)}{k\left( \sin C + \sin A \right)}\]
\[ = \frac{\left[ 2\cos\left( \frac{C + A}{2} \right)\cos\left( \frac{C - A}{2} \right) \right]\left[ - 2\sin\left( \frac{C + A}{2} \right)\sin\left( \frac{C - A}{2} \right) \right]}{2k\left( \sin C + \sin A \right)k}\]
\[ = \frac{- 2\cos\left( \frac{C + A}{2} \right)\cos\left( \frac{C - A}{2} \right)}{k}$=$\frac{- \left( \sin C - \sin A \right)}{k}=\frac{\sin A - \sin C}{k}\]
Similarly,
\[\frac{\cos^2 A - \cos^2 B}{a + b} = \frac{\sin B - \sin A}{k}\]
Thus,
\[LHS = \frac{\sin A - \sin C}{k} + \frac{\sin B - \sin A}{k} + \frac{\sin C - \sin B}{k}\]
\[ = 0 = RHS\]
Hence, in any triangle ABC,
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