In any ΔABC, prove that

Need to prove: $\frac{\left(\cos ^{2} \mathrm{~B}-\cos ^{2} \mathrm{C}\right)}{\mathrm{b}+\mathrm{c}}+\frac{\left(\cos ^{2} \mathrm{C}-\cos ^{2} \mathrm{~A}\right)}{\mathrm{c}+\mathrm{a}}+\frac{\left(\cos ^{2} \mathrm{~A}-\cos ^{2} \mathrm{~B}\right)}{\mathrm{a}+\mathrm{b}}=0$

We know that, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$ where $R$ is the circumradius.

Therefore,

a = 2R sinA ---- (a)

Similarly, b = 2R sinB and c = 2R sinC

From left hand side,

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

$=\frac{\left(1-\sin ^{2} \mathrm{~B}-1+\sin ^{2} \mathrm{C}\right)}{\mathrm{b}+\mathrm{c}}+\frac{\left(1-\sin ^{2} \mathrm{C}-1+\sin ^{2} \mathrm{~A}\right)}{\mathrm{c}+\mathrm{a}}$

$+\frac{\left(1-\sin ^{2} \mathrm{~A}-1+\sin ^{2} \mathrm{~B}\right)}{\mathrm{a}+\mathrm{b}}$

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

Now,

$=\frac{1}{2 R}\left[\frac{(\sin B+\sin C)(\sin C-\sin B)}{\sin B+\sin C}+\frac{(\sin A+\sin C)(\sin A-\sin C)}{\sin A+\sin C}\right.$

$\left.+\frac{(\sin A+\sin B)(\sin B-\sin A)}{\sin A+\sin B}\right]$

$=\frac{1}{2 R}[\sin C-\sin B+\sin A-\sin C+\sin B-\sin A]$

$=0$ [Proved]