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AP EAMCET · Maths · Properties of Triangles

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

  1. A \(a^2-b^2\)
  2. B \(\frac{1}{a^2}-\frac{1}{b^2}\)
  3. C \(a^2+b^2\)
  4. D \(\frac{1}{a^2}+\frac{1}{b^2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{a^2}-\frac{1}{b^2}\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \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}{a^2}-\frac{1}{b^2}-2\left[\frac{\sin ^2 A}{a^2}-\frac{\sin ^2 B}{b^2}\right] \\ & =\frac{1}{a^2}-\frac{1}{b^2} \quad\left[\because \frac{\sin…

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