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 $\frac{\cos ^{2} B-\cos ^{2} C}{b+c}+\frac{\cos ^{2} C-\cos ^{2} A}{c+a}+\frac{\cos ^{2} A-\cos ^{2} B}{a+b}=0$.

$\mathrm{LHS}=\frac{\cos ^{2} B-\cos ^{2} C}{b+c}+\frac{\cos ^{2} C-\cos ^{2} A}{c+a}+\frac{\cos ^{2} A-\cos ^{2} B}{a+b}$

Now,

$\frac{\cos ^{2} B-\cos ^{2} C}{b+c}=\frac{\cos ^{2} B-\cos ^{2} C}{k(\sin B+\sin C)}$

$=\frac{(\cos B+\cos C)(\cos B-\cos C)}{k(\sin B+\sin C)} \quad\left(\because \cos ^{2} B-\cos ^{2} C=(\cos B+\cos C)(\cos B-\cos C)\right)$

$=\frac{\left[2 \cos \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)\right]\left[-2 \sin \left(\frac{B+C}{2}\right) \sin \left(\frac{B-C}{2}\right)\right]}{2 k \sin \left(\frac{B+C}{2}\right) \sin \left(\frac{B-C}{2}\right)}$

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

Also,

$\frac{\cos ^{2} C-\cos ^{2} A}{c+a}=\frac{\cos ^{2} C-\cos ^{2} A}{k(\sin C+\sin A)}$

$=\frac{(\cos C+\cos A)(\cos C-\cos A)}{k(\sin C+\sin A)}$

$=\frac{\left[2 \cos \left(\frac{C+A}{2}\right) \cos \left(\frac{C-A}{2}\right)\right]\left[-2 \sin \left(\frac{C+A}{2}\right) \sin \left(\frac{C-A}{2}\right)\right]}{2 k(\sin C+\sin A) k}$

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

Similarly,

$\frac{\cos ^{2} A-\cos ^{2} B}{a+b}=\frac{\sin B-\sin A}{k}$

Thus,

LHS $=\frac{\sin A-\sin C}{k}+\frac{\sin B-\sin A}{k}+\frac{\sin C-\sin B}{k}$

$=0=\mathrm{RHS}$

Hence, in any triangle $A B C, \frac{\cos ^{2} B-\cos ^{2} C}{b+c}+\frac{\cos ^{2} C-\cos ^{2} A}{c+a}+\frac{\cos ^{2} A-\cos ^{2} B}{a+b}=0$.