In any $\triangle A B C, \frac{\cos 2 A}{a^2}-\frac{\cos 2 B}{b^2}=$

In any $\triangle A B C, \frac{\cos 2 A}{a^2}-\frac{\cos 2 B}{b^2}=$
  1. $a^2-b^2$
  2. $\frac{1}{a^2}-\frac{1}{b^2}$
  3. $a^2+b^2$
  4. $\frac{1}{a^2}+\frac{1}{b^2}$

Solution

$\begin{aligned} & \frac{\cos 2 A}{a^2}-\frac{\cos 2 B}{b^2}=\frac{1-2 \sin ^2 A}{a^2}-\frac{1-2 \sin ^2 B}{b^2} \\ & =\frac{1}{a^2}-\frac{1}{b^2}-2\left[\frac{\sin ^2 A}{a^2}-\frac{\sin ^2 B}{b^2}\right] \\ & =\frac{1}{a^2}-\frac{1}{b^2} \quad\left[\because \frac{\sin A}{a}=\frac{\sin B}{b}\right]\end{aligned}$

Asked in: AP EAMCET 2022 (07 Jul Shift 2)

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