Question

In any $△ABC$, prove the following: $\frac{\cos 2A}{a^2}-\frac{\cos 2B}{b^2}=(\frac{1}{a^2}-\frac{1}{b^2})$

Answer 1

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