Question

In triangle ABC, prove the following:

$\frac{\cos 2A}{a^2}-\frac{\cos 2B}{b^2}-\frac{1}{a^2}-\frac{1}{b^2}$

Answer 1

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Then,

Consider the LHS of the equation $\frac{\cos 2A}{a^2}-\frac{\cos 2B}{b^2}-\frac{1}{a^2}-\frac{1}{b^2}$.

$$\begin{gathered} \mathrm{LHS}=\frac{\cos 2A}{a^2}-\frac{\cos 2B}{b^2} \\ =\frac{1-2{\sin }^{2}A}{a^2}-\frac{1-2{\sin }^{2}B}{b^2} \\ =\frac{1-2{\displaystyle \frac{a^2}{k^2}}}{a^2}-\frac{1-2{\displaystyle \frac{b^2}{k^2}}}{b^2} \\ =\frac{{\displaystyle \frac{k^2-2a^2}{k^2}}}{a^2}-\frac{{\displaystyle \frac{k^2-2b^2}{k^2}}}{b^2} \\ =\frac{k^2{b}^2-2a^2{b}^2-k^2{a}^2+2a^2{b}^2}{a^2{b}^2} \\ =\frac{k^2\left(b^2-a^2\right)}{k^2{a}^2{b}^2} \\ =\frac{1}{a^2}-\frac{1}{b^2}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$