For positive acute angles $A$ and $B$, it is known that $\sin A=\frac{4}{5}$ and $\tan B=\frac{21}{20}$. Find the value of $\sin (A-B)$ in simplest form. Answer Attempt 1 out of 2

Step 1 ：Understand the problem: We are given the values of \(\sin A = \frac{4}{5}\) and \(\tan B = \frac{21}{20}\), where A and B are positive acute angles. We are asked to find the value of \(\sin (A-B)\).

Step 2 ：Use the formula for \(\sin (A-B)\), which is \(\sin A \cos B - \cos A \sin B\).

Step 3 ：Find the values of \(\cos A\) and \(\cos B\). Using the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\), we find that \(\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}\). Similarly, using the identity \(\tan B = \frac{\sin B}{\cos B}\) and the Pythagorean identity, we find that \(\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \left(\frac{21}{20}\right)^2 / \left(1 + \left(\frac{21}{20}\right)^2\right)} = \frac{20}{29}\).

Step 4 ：Substitute the values into the formula to get \(\sin (A-B) = \left(\frac{4}{5}\right) * \left(\frac{20}{29}\right) - \left(\frac{3}{5}\right) * \left(\frac{21}{29}\right) = \frac{80}{145} - \frac{63}{145} = \frac{17}{145}\).

Step 5 ：Check the result: The result is a valid value for the sine of an angle, so it meets the requirements of the problem.

Step 6 ：\(\boxed{\sin (A-B) = \frac{17}{145}}\)