Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively, then, $\frac{\cos 2A}{a^2}-\frac{\cos 2B}{b^2}=?$

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

Answer 1

The correct option is A $\frac{1}{a^2}-\frac{1}{b^2}$

$\frac{\cos 2A}{a^2}-\frac{\cos 2B}{B^2}=\left(\frac{2{\cos }^{2}A-1}{a^2}\right)-\left(\frac{2{\cos }^{2}B-1}{b^2}\right)=2(\frac{{\cos }^{2}A}{a^2}-\frac{{\cos }^{2}B}{b^2})+(\frac{1}{b^2}-\frac{1}{a^2})=2\{{\left(\frac{b^2+c^2-a^2}{2bc}\right)}^2\frac{1}{a^2}-{\left(\frac{a^2+c^2-b^2}{2ac}\right)}^2\frac{1}{b^2}\}+(\frac{1}{b^2}-\frac{1}{a^2})=\frac{1}{2}\left(\frac{{(b^2+c^2-a^2)}^2-{(a^2+c^2-b^2)}^2}{a^2{b}^2{c}^2}\right)+(\frac{1}{b^2}-\frac{1}{a^2})=\frac{2b^2{c}^2-2a^2{c}^2}{a^2{b}^2{c}^2}+(\frac{1}{b^2}-\frac{1}{a^2})=\frac{b^2-a^2}{a^2{b}^2}+(\frac{1}{b^2}-\frac{1}{a^2})=\frac{b^2-a^2}{a^2{b}^2}=\frac{1}{a^2}-\frac{1}{b^2}$

So option $A$ is correct.