In triangle ABC, prove the following:

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Then,

Consider the LHS of the equation $a^{2}\left(\cos ^{2} B-\cos ^{2} C\right)+b^{2}\left(\cos ^{2} C-\cos ^{2} A\right)+c^{2}\left(\cos ^{2} A-\cos ^{2} B\right)=0$.

LHS $=a^{2}\left(\cos ^{2} B-\cos ^{2} C\right)+b^{2}\left(\cos ^{2} C-\cos ^{2} A\right)+c^{2}\left(\cos ^{2} A-\cos ^{2} B\right)$

$=k^{2} \sin ^{2} A\left(1-\sin ^{2} B-1+\sin ^{2} C\right)+k^{2} \sin ^{2} B\left(1-\sin ^{2} C-1+\sin ^{2} A\right)+k^{2} \sin ^{2} C\left(1-\sin ^{2} A-1+\sin ^{2} B\right)$

$=k^{2} \sin ^{2} A\left(\sin ^{2} C-\sin ^{2} B\right)+k^{2} \sin ^{2} B\left(\sin ^{2} A-\sin ^{2} C\right)+k^{2} \sin ^{2} C\left(\sin ^{2} B-\sin ^{2} A\right)$

$=k^{2}\left(\sin ^{2} A \sin ^{2} C-\sin ^{2} A \sin ^{2} B+\sin ^{2} A \sin ^{2} B-\sin ^{2} B \sin ^{2} C+\sin ^{2} C \sin ^{2} B-\sin ^{2} C \sin ^{2} A\right)$

$=k^{2} \times 0=0=$ RHS

Hence proved.