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

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

\[LHS = \frac{\cos2A}{a^2} - \frac{\cos2B}{b^2}\]
\[ = \frac{1 - 2 \sin^2 A}{a^2} - \frac{1 - 2 \sin^2 B}{b^2} \]
\[ = \frac{1 - 2\frac{a^2}{k^2}}{a^2} - \frac{1 - 2\frac{b^2}{k^2}}{b^2} \]
\[ = \frac{\frac{k^2 - 2 a^2}{k^2}}{a^2} - \frac{\frac{k^2 - 2 b^2}{k^2}}{b^2}\]
\[ = \frac{k^2 b^2 - 2 a^2 b^2 - k^2 a^2 + 2 a^2 b^2}{a^2 b^2}\]
\[ = \frac{k^2 \left( b^2 - a^2 \right)}{k^2 a^2 b^2}\]
\[ = \frac{1}{a^2} - \frac{1}{b^2} = RHS\]
\[\text { Hence proved } .\]