Question

In any triangle \(ABC\), \[ \frac{\cos 2A}{a^2}-\frac{\cos 2B}{b^2} = \] is equal to:

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

Hint:

For triangle problems involving \(\cos 2A\) and sides, the identity \[ \frac{\cos 2A}{a^2} = \frac{1}{a^2}-\frac{1}{2R^2} \] is extremely useful and helps avoid lengthy algebraic manipulations.

Answer 1

The Correct Option is B

Step 1: Use the cosine rule.
In a triangle, \[ \cos A=\frac{b^2+c^2-a^2}{2bc} \] and \[ \cos B=\frac{a^2+c^2-b^2}{2ac} \] Using \[ \cos 2A=2\cos^2A-1 \] and \[ \cos 2B=2\cos^2B-1 \] direct simplification becomes lengthy. Hence we use a standard identity valid in any triangle: \[ \frac{\cos 2A}{a^2} = \frac{1}{a^2} -\frac{1}{2R^2} \] and \[ \frac{\cos 2B}{b^2} = \frac{1}{b^2} -\frac{1}{2R^2} \] where \(R\) is the circumradius.

Step 2: Subtract the two expressions.
Therefore, \[ \frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2} = \left(\frac{1}{a^2}-\frac{1}{2R^2}\right) - \left(\frac{1}{b^2}-\frac{1}{2R^2}\right) \] \[ = \frac{1}{a^2} - \frac{1}{b^2} \]

Step 3: Final conclusion.
Hence, \[ \boxed{\frac{1}{a^2}-\frac{1}{b^2}} \] Therefore, the correct option is \[ \boxed{(2)\ \frac{1}{a^2}-\frac{1}{b^2}} \]