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 $\sin s=\sqrt{1-{\cos }^{2}s}=\sqrt{1-{(-\frac{3}{5})}^2}=\frac{4}{5}$.

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 $\cos t=-\sqrt{1-{\sin }^{2}t}=-\sqrt{1-{(-\frac{12}{13})}^2}=-\frac{5}{13}$.

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) $\sin (s+1)=\boxed{{\displaystyle 0.86}}$, (b) $\tan (s+1)=\boxed{{\displaystyle 0.57}}$, (c) The quadrant of $s+1$ is $\boxed{{\displaystyle \text{Quadrant III}}}$