In triangle ABC, prove the following:
Question:

In triangle ABC, prove the following:

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

Solution:

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

Then,

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

$\mathrm{LHS}=\frac{\cos 2 A}{a^{2}}-\frac{\cos 2 B}{b^{2}}$

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

$=\frac{1-2 \frac{a^{2}}{k^{2}}}{a^{2}}-\frac{1-2 \frac{b^{2}}{k^{2}}}{b^{2}}$

$=\frac{\frac{k^{2}-2 a^{2}}{k^{2}}}{a^{2}}-\frac{\frac{k^{2}-2 b^{2}}{k^{2}}}{b^{2}}$

$=\frac{k^{2} b^{2}-2 a^{2} b^{2}-k^{2} a^{2}+2 a^{2} b^{2}}{a^{2} b^{2}}$

$=\frac{k^{2}\left(b^{2}-a^{2}\right)}{k^{2} a^{2} b^{2}}$

$=\frac{1}{a^{2}}-\frac{1}{b^{2}}=\mathrm{RHS}$

Hence proved.

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