For positive acute angles $A$ and $B$, it is known that $\cos A=\frac{21}{29}$ and $\tan B=\frac{15}{8}$. Find the value of $\sin (A-B)$ in simplest form.

Step 1 ：Given: \(\cos A = \frac{21}{29}\) and \(\tan B = \frac{15}{8}\)

Step 2 ：Find: \(\sin (A-B)\)

Step 3 ：Use the formula: \(\sin (A-B) = \sin A \cos B - \cos A \sin B\)

Step 4 ：Find \(\sin A\) using \(\sin^2 A + \cos^2 A = 1\): \(\sin A = \frac{20}{29}\)

Step 5 ：Find \(\sin B\) and \(\cos B\) using \(\tan B = \frac{\sin B}{\cos B}\) and \(\sin^2 B + \cos^2 B = 1\): \(\sin B = \frac{15}{17}\) and \(\cos B = \frac{8}{17}\)

Step 6 ：Substitute the values into the formula: \(\sin (A-B) = \frac{20}{29} \cdot \frac{8}{17} - \frac{21}{29} \cdot \frac{15}{17}\)

Step 7 ：Simplify: \(\sin (A-B) = \boxed{-\frac{99}{493}}\)