Use the given information to find $(a) \sin (s+t),(b) \tan (s+t)$, and $(c)$ the quadrant of $s+t$. $\cos s=\frac{3}{5}$ and $\sin t=-\frac{12}{13}, s$ and $t$ in quadrant IV (a) $\sin (s+t)=$ (Use integers or fractions for any numbers in the expression.)

Step 1 ：We are given that \(\cos s = \frac{3}{5}\) and \(\sin t = -\frac{12}{13}\).

Step 2 ：We also know that both s and t are in quadrant IV, where \(\sin\) is negative and \(\cos\) is positive.

Step 3 ：Therefore, we can calculate \(\sin s = -\sqrt{1 - \cos^2 s}\) and \(\cos t = \sqrt{1 - \sin^2 t}\).

Step 4 ：Substituting the given values, we get \(\sin s = -\sqrt{1 - (\frac{3}{5})^2} = -0.8\) and \(\cos t = \sqrt{1 - (-\frac{12}{13})^2} = 0.38461538461538447\).

Step 5 ：We know that \(\sin (s+t) = \sin s \cos t + \cos s \sin t\).

Step 6 ：Substituting the calculated and given values, we get \(\sin (s+t) = -0.8 * 0.38461538461538447 + 0.6 * -0.9230769230769231 = -0.8615384615384615\).

Step 7 ：So, the final answer is \(\sin (s+t) = \boxed{-0.8615384615384615}\).