KCET · Maths · Properties of Triangles
In any \(\triangle A B C\), the simplified form of \(\frac{\cos 2 A}{a^{2}}-\frac{\cos 2 B}{b^{2}}\) is
- A \(a^{2}-b^{2}\)
- B \(\frac{1}{a^{2}-b^{2}}\)
- C \(\frac{1}{a^{2}}-\frac{1}{b^{2}}\)
- D \(a^{2}+b^{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{a^{2}}-\frac{1}{b^{2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \frac{\cos 2 A}{a^{2}} &-\frac{\cos 2 B}{b^{2}} \\ &=\frac{\left(1-2 \sin ^{2} A\right)}{a^{2}}-\frac{\left(1-2 \sin ^{2} B\right)}{b^{2}} \\ &=\frac{\left(1-2 a^{2} k^{2}\right)}{a^{2}}-\frac{\left(1-2 b^{2} k^{2}\right)}{b^{2}} \\ &=\left(\frac{1}{a^{2}}-2 k^{2}\right)-\left(\frac{1}{b^{2}}-2 k^{2}\right) \\ &=\left(\frac{1}{a^{2}}-\frac{1}{b^{2}}\right)-2 k^{2}+2 k^{2} \\ &=\frac{1}{a^{2}}-\frac{1}{b^{2}} \end{aligned}\)
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