Question

In $\Delta ABC,\frac{\cos 2A}{a^2}-\frac{\cos 2B}{b^2}$ is equal to:

• A. $c^2/a^2{b}^2$
• B. $\frac{1}{a^2}-\frac{1}{b^2}$
• C. $\frac{1}{ab}$
• D. none of these

Answer 1

Answer: (B)

$\therefore$ $\frac{\cos 2A}{a^2}-\frac{\cos 2B}{b^2}$
$=\frac{1-2{\sin }^{2}A}{a^2}-\frac{1-2{\sin }^{2}B}{b^2}$
Applying sine rule
$=\frac{1}{a^2}-\frac{1}{b^2}-2k(1-1)$
$=\frac{1}{a^2}-\frac{1}{b^2}$