Answer
(a) $$\sin(s+t)=-\frac{63}{65}$$
(b) $$\tan(s+t)=-\frac{63}{16}$$
(c) $(s+t)$ lies in quadrant IV.
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 $\tan t$.
$$\sin s=\frac{3}{5}\hspace{1.5cm}\sin t=-\frac{12}{13}\hspace{1.5cm}\text{s in quadrant I and t in quadrant III}$$
1) Find $\sin s$, $\sin t$, $\cos s$, $\cos t$, $\tan s$ and $\tan t$.
$s$ is in quadrant I, so all $\sin s$, $\cos s$ and $\tan s$ are positive.
$t$ is in quadrant III, so $\sin t\lt0$, $\cos t\lt0$ and $\tan t=\frac{\sin t}{\cos t}\gt0$
$\cos^2 s=1-\sin^2 s=1-\Big(\frac{3}{5}\Big)^2=1-\frac{9}{25}=\frac{16}{25}$
$\cos s=\frac{4}{5}$
$\cos^2 t=1-\sin^2t=1-\Big(-\frac{12}{13}\Big)^2=1-\frac{144}{169}=\frac{25}{169}$
$\cos t=-\frac{5}{13}$
$$\tan s=\frac{\sin s}{\cos s}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$$
$$\tan t=\frac{\sin t}{\cos t}=\frac{-\frac{12}{13}}{-\frac{5}{13}}=\frac{12}{5}$$
(a) Find $\sin(s+t)$
$$\sin(s+t)=\sin s\cos t+\cos s\sin t\hspace{1.5cm}\text{(sine sum identity)}$$
$$\sin(s+t)=\frac{3}{5}\times\Big(-\frac{5}{13}\Big)+\frac{4}{5}\times\Big(-\frac{12}{13}\Big)$$
$$\sin(s+t)=-\frac{15}{65}-\frac{48}{65}$$
$$\sin(s+t)=-\frac{63}{65}$$
(b) Find $\tan(s+t)$
$$\tan(s+t)=\frac{\tan s+\tan t}{1-\tan s\tan t}\hspace{1.5cm}\text{(tangent sum identity)}$$
$$\tan(s+t)=\frac{\frac{3}{4}+\frac{12}{5}}{1-\frac{3}{4}\times\frac{12}{5}}$$
$$\tan(s+t)=\frac{\frac{63}{20}}{1-\frac{9}{5}}$$
$$\tan(s+t)=\frac{\frac{63}{20}}{-\frac{4}{5}}=-\frac{63\times5}{20\times4}$$
$$\tan(s+t)=-\frac{63}{16}$$
(c) We see that $\sin(s+t)\lt0$ and $\tan (s+t)\lt0$.
$\tan (s+t)=\frac{\sin(s+t)}{\cos(s+t)}$, this means $\cos(s+t)\gt0$
$\sin(s+t)\lt0$ but $\cos(s+t)\gt0$ show that $(s+t)$ lies in quadrant IV.
