Question:

<p><strong>Step 1:</strong> Find $\sin(a+b)$ from $\cos(a+b)=\frac{3}{5}$:</p><p>Since $0 < a,b < \frac{\pi}{4}$, we have $0 < a+b < \frac{\pi}{2}$, so $\sin(a+b) > 0$</p><p>$\sin(a+b) = \sqrt{1-\cos^2(a+b)} = \sqrt{1-\frac{9}{25}} = \frac{4}{5}$</p><p><strong>Step 2:</strong> Find $\cos(a-b)$ from $\sin(a-b)=\frac{5}{13}$:</p><p>Since $0 < a,b < \frac{\pi}{4}$, we have $-\frac{\pi}{4} < a-b < \frac{\pi}{4}$, so $\cos(a-b) > 0$</p><p>$\cos(a-b) = \sqrt{1-\sin^2(a-b)} = \sqrt{1-\frac{25}{169}} = \frac{12}{13}$</p><p><strong>Step 3:</strong> Use $2a = (a+b)+(a-b)$:</p><p>$\sin(2a) = \sin(a+b)\cos(a-b) + \cos(a+b)\sin(a-b) = \frac{4}{5}\cdot\frac{12}{13} + \frac{3}{5}\cdot\frac{5}{13} = \frac{48}{65}+\frac{15}{65} = \frac{63}{65}$</p><p>$\cos(2a) = \cos(a+b)\cos(a-b) - \sin(a+b)\sin(a-b) = \frac{3}{5}\cdot\frac{12}{13} - \frac{4}{5}\cdot\frac{5}{13} = \frac{36}{65}-\frac{20}{65} = \frac{16}{65}$</p><p><strong>Step 4:</strong> $\tan(2a) = \frac{\sin(2a)}{\cos(2a)} = \frac{\frac{63}{65}}{\frac{16}{65}} = \frac{63}{16}$</p><p>∴ Answer is (b) $\frac{63}{16}$</p>