AP EAMCET · Maths · Properties of Triangles
In any \(\triangle A B C, \frac{\cos 2 A}{a^2}-\frac{\cos 2 B}{b^2}=\)
- A \(a^2-b^2\)
- B \(\frac{1}{a^2}-\frac{1}{b^2}\)
- C \(a^2+b^2\)
- D \(\frac{1}{a^2}+\frac{1}{b^2}\)
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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