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In a triangle $\mathrm{ABC}$ with usual notations $\mathrm{a}=2, \mathrm{~b}=3$, then value of $\frac{\cos 2 A}{a^2}-\frac{\cos 2 B}{b^2}$ is
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Verified Answer
The correct answer is:
$\frac{5}{36}$
$\begin{aligned}
& \frac{\cos 2 \mathrm{~A}}{\mathrm{a}^2}-\frac{\cos 2 \mathrm{~B}}{\mathrm{~b}^2} \\
& =\frac{1-2 \sin ^2 \mathrm{~A}}{\mathrm{a}^2}-\frac{1-2 \sin ^2 \mathrm{~B}}{\mathrm{~b}^2}=\left(\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\right)-2\left(\frac{\sin ^2 \mathrm{~A}}{\mathrm{a}^2}-\frac{\sin ^2 \mathrm{~B}}{\mathrm{~b}^2}\right)
\end{aligned}$
From sine rule, we know that
$\begin{aligned}
& \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}} \\
& \therefore \text { Given Expression }=\left(\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\right)-0=\frac{1}{2^2}-\frac{1}{3^2}=\frac{1}{4}-\frac{1}{9}=\frac{5}{36}
\end{aligned}$
& \frac{\cos 2 \mathrm{~A}}{\mathrm{a}^2}-\frac{\cos 2 \mathrm{~B}}{\mathrm{~b}^2} \\
& =\frac{1-2 \sin ^2 \mathrm{~A}}{\mathrm{a}^2}-\frac{1-2 \sin ^2 \mathrm{~B}}{\mathrm{~b}^2}=\left(\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\right)-2\left(\frac{\sin ^2 \mathrm{~A}}{\mathrm{a}^2}-\frac{\sin ^2 \mathrm{~B}}{\mathrm{~b}^2}\right)
\end{aligned}$
From sine rule, we know that
$\begin{aligned}
& \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}} \\
& \therefore \text { Given Expression }=\left(\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\right)-0=\frac{1}{2^2}-\frac{1}{3^2}=\frac{1}{4}-\frac{1}{9}=\frac{5}{36}
\end{aligned}$
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