In a triangle ABC with usual notations a=2 an | vedclass

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(A) Using the identity $\cos 2	heta = 1 - 2\sin^2	heta$,the expression becomes: $\frac{1 - 2\sin^2 A}{a^2} - \frac{1 - 2\sin^2 B}{b^2} = \left(\frac

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$\frac{5}{36}$

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(A) Using the identity $\cos 2	heta = 1 - 2\sin^2	heta$,the expression becomes: $\frac{1 - 2\sin^2 A}{a^2} - \frac{1 - 2\sin^2 B}{b^2} = \left(\frac{1}{a^2} - \frac{1}{b^2}ight) - 2\left(\frac{\sin^2 A}{a^2} - \frac{\sin^2 B}{b^2}ight)$ From the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = k$,which implies $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{1}{k}$. Therefore,$\frac{\sin^2 A}{a^2} = \frac{\sin^2 B}{b^2} = \frac{1}{k^2}$. Substituting this into the expression: $\left(\frac{1}{a^2} - \frac{1}{b^2}ight) - 2\left(\frac{1}{k^2} - \frac{1}{k^2}ight) = \frac{1}{a^2} - \frac{1}{b^2}$ Given $a=2$ and $b=3$: $\frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$

In a triangle $ABC$ with usual notations $a=2$ and $b=3$,the value of $\frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2}$ is:

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