Question

In any $△ABC,\frac{\cos 2A}{a^2}-\frac{\cos 2B}{b^2}$ is independent of the measures of the angles $A$ and $B$

• A. True
• B. False

Answer 1

The correct option is B False

$\frac{\cos 2A}{a^2}-\frac{\cos 2B}{b^2}$
$=\frac{1-2{\sin }^{2}A}{a^2}-\frac{1-2{\sin }^{2}B}{b^2}$
$=\frac{1}{a^2}-\frac{1}{b^2}-2[\frac{{\sin }^{2}A}{a^2}-\frac{{\sin }^{2}B}{b^2}]$
Since $\frac{\sin A}{a}=\frac{\sin B}{b}$ we have
$=\frac{1}{a^2}-\frac{1}{b^2}-2[\frac{1}{a^2}-\frac{1}{b^2}]$
$=\frac{1}{b^2}-\frac{1}{a^2}$ which is not independent of the measure of $A$ and $B$.