Question

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

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