Question
In $\triangle A B C$ prove that $a^2\left(\cos ^2 B-\cos ^2 C\right)+b^2\left(\cos ^2 C-\cos ^2 A\right)+c^2\left(\cos ^2 A-\cos ^2 B\right)=0$.
$\begin{aligned} & \frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}=k \\ & \text { LHS }=a^2\left(\cos ^2 B-\cos ^2 C\right)+b^2\left(\cos ^2 C-\cos ^2 A\right)+c^2\left(\cos ^2 A-\cos ^2 B\right)\end{aligned}$
$\begin{aligned} & =k^2 \sin ^2 A\left[\left(1-\sin ^2 B\right)-\left(1-\sin ^2 C\right)\right]+k^2 \sin ^2 B\left[\left(1-\sin ^2 C\right)-\left(1-\sin ^2 A\right)\right]+k^2 \sin ^2 C\left[\left(1-\sin ^2 A\right)\right. \\ & \left.-\left(1-\sin ^2 B\right)\right]\end{aligned}$
$\begin{aligned} & =k^2 \sin ^2 A\left(\sin ^2 C-\sin ^2 B\right)+k^2 \sin ^2 B\left(\sin ^2 A-\sin ^2 C\right)+k^2 \sin ^2 C\left(\sin ^2 B-\sin ^2 A\right) \\ & =k^2\left(\sin ^2 A \sin ^2 C-\sin ^2 A \sin ^2 B+\sin ^2 A \sin ^2 B-\sin ^2 B \sin ^2 C+\sin ^2 B \sin ^2 C-\sin ^2 A \sin ^2 C\right) \\ & =k^2(0)=0=\text { RHS. }\end{aligned}$
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