Question

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

Answer 1

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

$\frac{\sin A}{a}=\frac{\sin B}{b}$

$\Rightarrow b\sin A=a\sin B$

$\Rightarrow b^2\cos 2A-a^2\cos 2B-a^2\cos 2B=b^2-2a^2{\sin }^{2}B-a^2+2a^2{\sin }^{2}B$

$\therefore b^2\cos 2A-a^2\cos 2B=b^2-a^2$

LHS=RHS

Hence proved.