# In any ΔABC, prove that

Question:

In any ΔABC, prove that

$\frac{\cos 2 A}{a^{2}}-\frac{\cos 2 B}{b^{2}}=\left(\frac{1}{a^{2}}-\frac{1}{b^{2}}\right)$

Solution:

Need to prove: $\frac{\cos 2 \mathrm{~A}}{\mathrm{a}^{2}}-\frac{\cos 2 \mathrm{~B}}{\mathrm{~b}^{2}}=\left(\frac{1}{\mathrm{a}^{2}}-\frac{1}{\mathrm{~b}^{2}}\right)$

Left hand side,

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

$=\frac{1-2 \sin ^{2} A}{a^{2}}-\frac{1-2 \sin ^{2} B}{b^{2}}$

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

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,

$\frac{\sin ^{2} \mathrm{~B}}{\mathrm{~b}^{2}}-\frac{\sin ^{2} \mathrm{~A}}{\mathrm{a}^{2}}=\frac{1}{4 \mathrm{R}^{2}}-\frac{1}{4 \mathrm{R}^{2}}=0$

Hence,

$\frac{\cos 2 \mathrm{~A}}{\mathrm{a}^{2}}-\frac{\cos 2 \mathrm{~B}}{\mathrm{~b}^{2}}=\frac{1}{\mathrm{a}^{2}}-\frac{1}{\mathrm{~b}^{2}}$ [Proved]