Answer
(a) $$\sin(s+t)=-\frac{8\sqrt6+3}{25}$$
(b) $$\tan(s+t)=-\frac{8\sqrt6+3}{4-6\sqrt6}$$
(c) $(s+t)$ is in quadrant III.
Work Step by Step
To find $\sin(s+t)$ and $\tan(s+t)$, all values of $\sin s$, $\sin t$, $\cos s$, $\cos t$, $\tan s$ and $\tan t$ must be known.
Therefore, the privileged job is to find all 6 essential values of $\sin s$, $\sin t$, $\cos s$, $\cos t$, $\tan s$ and $\tan t$.
$$\cos s=-\frac{1}{5}\hspace{1.5cm}\sin t=\frac{3}{5}\hspace{1.5cm}\text{s and t are in quadrant II}$$
1) Find $\sin s$, $\sin t$, $\cos s$, $\cos t$, $\tan s$ and $\tan t$.
$s$ and $t$ are in quadrant II, so $\sin s\gt0$, $\sin t\gt0$ but $\cos s\lt0$ and $\cos t\lt0$.
That means $\tan s=\frac{\sin s}{\cos s}\lt0$ and $\tan t=\frac{\sin t}{\cos t}\lt0$
$\sin^2 s=1-\cos^2 s=1-\Big(-\frac{1}{5}\Big)^2=1-\frac{1}{25}=\frac{24}{25}$
$\sin s=\frac{\sqrt{24}}{5}=\frac{2\sqrt6}{5}$
$\cos^2 t=1-\sin^2t=1-\Big(\frac{3}{5}\Big)^2=1-\frac{9}{25}=\frac{16}{25}$
$\cos t=-\frac{4}{5}$
$$\tan s=\frac{\sin s}{\cos s}=\frac{\frac{2\sqrt6}{5}}{-\frac{1}{5}}=-\frac{2\sqrt6}{1}=-2\sqrt6$$
$$\tan t=\frac{\sin t}{\cos t}=\frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4}$$
(a) Find $\sin(s+t)$
$$\sin(s+t)=\sin s\cos t+\cos s\sin t\hspace{1.5cm}\text{(sine sum identity)}$$
$$\sin(s+t)=\frac{2\sqrt6}{5}\times\Big(-\frac{4}{5}\Big)+\Big(-\frac{1}{5}\Big)\times\frac{3}{5}$$
$$\sin(s+t)=-\frac{8\sqrt6}{25}-\frac{3}{25}$$
$$\sin(s+t)=\frac{-8\sqrt6-3}{25}=-\frac{8\sqrt6+3}{25}$$
(b) Find $\tan(s+t)$
$$\tan(s+t)=\frac{\tan s+\tan t}{1-\tan s\tan t}\hspace{1.5cm}\text{(tangent sum identity)}$$
$$\tan(s+t)=\frac{-2\sqrt6-\frac{3}{4}}{1-(-2\sqrt6)\times\Big(-\frac{3}{4}\Big)}$$
$$\tan(s+t)=\frac{\frac{-8\sqrt6-3}{4}}{1-\frac{3\sqrt6}{2}}$$
$$\tan(s+t)=\frac{\frac{-8\sqrt6-3}{4}}{\frac{2-3\sqrt6}{2}}=\frac{(-8\sqrt6-3)\times2}{(2-3\sqrt6)\times4}$$
$$\tan(s+t)=\frac{-8\sqrt6-3}{2(2-3\sqrt6)}=-\frac{8\sqrt6+3}{4-6\sqrt6}$$
(c) We see that $\sin(s+t)\lt0$.
$8\sqrt6+3\gt0$, and since $4\lt6\sqrt6$, $4-6\sqrt{6}\lt0$, so $\frac{8\sqrt6+3}{4-6\sqrt6}\lt0$. Therefore, $\tan(s+t)=-\frac{8\sqrt6+3}{4-6\sqrt6}\gt0$
$\tan (s+t)=\frac{\sin(s+t)}{\cos(s+t)}$, this means $\cos(s+t)\lt0$
As $\sin(s+t)\lt0$ and $\cos(s+t)\lt0$, $(s+t)$ would in quadrant III.
