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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Aicharya · Accepted Answer

$\begin{aligned} & \frac{\cos 2 \mathrm{~A}}{\mathrm{a}^2}-\frac{\cos 2 \mathrm{~B}}{\mathrm{~b}^2} \ & =\frac{1-2 \sin ^2 \mathrm{~A}}{\mathrm{a}^2}-\frac{1-2 \sin ^2 \mathrm{~B}}{\mathrm{~b}^2}=\left(\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}ight)-2\left(\frac{\sin ^2 \mathrm{~A}}{\mathrm{a}^2}-\frac{\sin ^2 \mathrm{~B}}{\mathrm{~b}^2}ight) \end{aligned}$ From sine rule, we know that $\begin{aligned} & \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}} \ & 	herefore 	ext { Given Expression }=\left(\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}ight)-0=\frac{1}{2^2}-\frac{1}{3^2}=\frac{1}{4}-\frac{1}{9}=\frac{5}{36} \end{aligned}$

In a triangle $\mathrm{ABC}$ with usual notations $\mathrm{a}=2, \mathrm{~b}=3$, then value of $\frac{\cos 2…

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