2A a^2 - 2B b^2 =1 a^2 -1 b^2

Question

$\frac{\cos2\text{A}}{\text{a}^2}-\frac{\cos2\text{B}}{\text{b}^2}=\frac{1}{\text{a}^2}-\frac{1}{\text{b}^2}$

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Answer

$\text{LHS}=\frac{\cos2\text{A}}{\text{a}^2}-\frac{\cos2\text{B}}{\text{b}^2}$ $\frac{\sin\text{A}}{\text{a}}=\frac{\sin\text{B}}{\text{b}}=\frac{\sin\text{C}}{\text{c}}=\text{k}$ $=\frac{1-2\sin^2\text{A}}{\text{a}^2}-\frac{1-2\sin^2\text{B}}{\text{b}^2}$ $=\frac{1}{\text{a}^2}-\frac{1}{\text{b}^2}-2\Big(\frac{\sin^2\text{A}}{\text{a}^2}-\frac{\sin^2\text{B}}{\text{b}^2}\Big)$ $=\frac{1}{\text{a}^2}-\frac{1}{\text{b}^2}-2(\text{k}^2-\text{k}^2)$ [Using sine rule] $=\frac{1}{\text{a}^2}-\frac{1}{\text{b}^2}=\text{RHS}$ Hence Proved

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