Use the given information to find (a) $\sin (s+1)$, (b) $\tan (s+1)$, and $(c)$ the quadrant of $s+t$.
\[
\cos s=-\frac{3}{5} \text { and } \sin t=-\frac{12}{13}, s \text { and } t \text { in quadrant int }
\]
(a) $\sin (s+t)=$
(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression )
(b) $\tan (s+t)=$
(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression )
(c) What is the quadrant of $s+1$ ?
Quadrant IV
Quadrant II
Quadrant III
Quadrant I

Step 1 ：Given that \(\cos s = -\frac{3}{5}\) and \(\sin t = -\frac{12}{13}\), we need to find the values of \(\sin s\) and \(\cos t\) to calculate \(\sin (s+1)\) and \(\tan (s+1)\).

Step 2 ：Using the Pythagorean identity \(\sin^2 s + \cos^2 s = 1\), we can find \(\sin s\). Since \(s\) is in quadrant II where sine is positive, we get \(\sin s = \sqrt{1 - \cos^2 s} = \sqrt{1 - \left(-\frac{3}{5}\right)^2} = \frac{4}{5}\).

Step 3 ：Similarly, using the Pythagorean identity \(\sin^2 t + \cos^2 t = 1\), we can find \(\cos t\). Since \(t\) is in quadrant III where cosine is negative, we get \(\cos t = -\sqrt{1 - \sin^2 t} = -\sqrt{1 - \left(-\frac{12}{13}\right)^2} = -\frac{5}{13}\).

Step 4 ：Using the addition formulas for sine and tangent, we can find \(\sin (s+1)\) and \(\tan (s+1)\).

Step 5 ：\(\sin (s+1) = \sin s \cos 1 + \cos s \sin 1 = \frac{4}{5} \cos 1 - \frac{3}{5} \sin 1\).

Step 6 ：\(\tan (s+1) = \frac{\sin (s+1)}{\cos (s+1)} = \frac{\frac{4}{5} \cos 1 - \frac{3}{5} \sin 1}{\cos s \cos 1 - \sin s \sin 1} = \frac{-\frac{3}{5} \cos 1 - \frac{4}{5} \sin 1}{-\frac{3}{5} \sin 1 + \frac{4}{5} \cos 1}\).

Step 7 ：The quadrant of \(s+1\) depends on the quadrant of \(s\). If \(s\) is in quadrant II, then \(s+1\) is in quadrant III. Since \(s\) is in quadrant II, \(s+1\) is in quadrant III.

Step 8 ：Final Answer: (a) \(\sin (s+1) = \boxed{0.86}\), (b) \(\tan (s+1) = \boxed{0.57}\), (c) The quadrant of \(s+1\) is \(\boxed{\text{Quadrant III}}\)