Question

In any triangle ABC, \( \frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2} = \)

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

Hint:

Use double angle formula: \(\cos 2\theta = 1 - 2\sin^2\theta\) (or \(2\cos^2\theta - 1\) or \(\cos^2\theta - \sin^2\theta\)).
Use Sine Rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\).
Substitute and simplify.

Answer 1

The Correct Option is B

We use the identity \(\cos 2\theta = 1 - 2\sin^2\theta\). So, \(\cos 2A = 1 - 2\sin^2 A\) and \(\cos 2B = 1 - 2\sin^2 B\). The expression is \( E = \frac{1 - 2\sin^2 A}{a^2} - \frac{1 - 2\sin^2 B}{b^2} \). \( E = \frac{1}{a^2} - \frac{2\sin^2 A}{a^2} - \frac{1}{b^2} + \frac{2\sin^2 B}{b^2} \). By the Sine Rule in \(\triangle ABC\), we have \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\), where R is the circumradius. So, \(\sin A = \frac{a}{2R}\) and \(\sin B = \frac{b}{2R}\). Substitute these into the expression for E: \(\frac{2\sin^2 A}{a^2} = \frac{2 (a/(2R))^2}{a^2} = \frac{2 (a^2/(4R^2))}{a^2} = \frac{2a^2}{4R^2 a^2} = \frac{2}{4R^2} = \frac{1}{2R^2}\). Similarly, \(\frac{2\sin^2 B}{b^2} = \frac{2 (b/(2R))^2}{b^2} = \frac{2 (b^2/(4R^2))}{b^2} = \frac{2b^2}{4R^2 b^2} = \frac{2}{4R^2} = \frac{1}{2R^2}\). So, the expression becomes: \( E = \frac{1}{a^2} - \frac{1}{2R^2} - \frac{1}{b^2} + \frac{1}{2R^2} \) \( E = \frac{1}{a^2} - \frac{1}{b^2} \). This matches option (b). \[ \boxed{\frac{1}{a^2} - \frac{1}{b^2}} \]