Question

In any triangle $ABC$, the simplified form of $\frac{Cos2A}{a^2}-\frac{Cos2B}{b^2}$ is

• A. $a^2-b^2$
• B. $\frac{1}{a^2-b^2}$
• C. $\frac{1}{a^2}-\frac{1}{b^2}$
• D. $a^2+b^2$

Answer 1

Answer: (C)

$\frac{\cos 2A}{a^2}-\frac{\cos 2B}{b^2}$
$=\frac{\left(1-2{\sin }^{2}A\right)}{a^2}-\frac{\left(1-2{\sin }^{2}B\right)}{b^2}$
$=\frac{\left(1-2a^2{k}^2\right)}{a^2}-\frac{\left(1-2b^2{k}^2\right)}{b^2}$
$\left(\because \frac{\sin A}{a}=\frac{\sin B}{b}=k\right)$
$=\left(\frac{1}{a^2}-2k^2\right)-\left(\frac{1}{b^2}-2k^2\right)$
$=\left(\frac{1}{a^2}-\frac{1}{b^2}\right)-2k^2+2k^2$
$=\frac{1}{a^2}-\frac{1}{b^2}$