# In any ΔABC, prove that

Question:

In any ΔABC, prove that

$a^{2}\left(\cos ^{2} B-\cos ^{2} C\right)+b^{2}\left(\cos ^{2} C-\cos ^{2} A\right)+c^{2}\left(\cos ^{2} A-\cos ^{2} B\right)=0$

Solution:

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

From left hand side,

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

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

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

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

Therefore

$a=2 R \sin A \cdots$ (a)

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

So,

$=4 R^{2}\left[\sin ^{2} A\left(-\sin ^{2} B+\sin ^{2} C\right)+\sin ^{2} B\left(-\sin ^{2} C+\sin ^{2} A\right)+\sin ^{2} C\left(-\sin ^{2} A+\sin ^{2} B\right)\right.$

$=4 R^{2}\left[-\sin ^{2} A \sin ^{2} B+\sin ^{2} A \sin ^{2} C-\sin ^{2} B \sin ^{2} C+\sin ^{2} A \sin ^{2} B-\sin ^{2} A \sin ^{2} C+\sin ^{2} B \sin ^{2} C\right]$

$=4 R^{2}[0]$

$=0$ [Proved]