Question

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

Answer 1

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

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

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

Using sine rule $\sin A=\frac{a}{2R},\sin B=\frac{b}{2R}$

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

$=\frac{1}{a^2}-\frac{1}{b^2}-2(\frac{a^2}{4R^2{a}^2}-\frac{b^2}{4R^2{b}^2})$

$=\frac{1}{a^2}-\frac{1}{b^2}-2(\frac{1}{4R^2}-\frac{1}{4R^2})$

$=\frac{1}{a^2}-\frac{1}{b^2}-2\left(0\right)$

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