Question

$\frac{cos2A}{a^2}-\frac{cos2B}{b^2}=\frac{1}{a^2}-\frac{1}{b^2}$

Answer 1

$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=kLHS=\frac{1-2si{n}^{2}A}{a^2}-\frac{1-2si{n}^{2}B}{b^2}=\frac{1}{a^2}-\frac{1}{b^2}-2(\frac{si{n}^{2}A}{a^2}-\frac{si{n}^{2}B}{b^2})=\frac{1}{a^2}-\frac{1}{b^2}-2(k^2-k^2)\text{[Using sine rule]}=\frac{1}{a^2}-\frac{1}{b^2}=RHS\text{Hence proved.}$