To solve the problem, let's break it into steps:

Given Information:

1. $\cos(s) = -\frac{3}{5}$, $s$ is in Quadrant III.
2. $\sin(t) = -\frac{12}{13}$, $t$ is in Quadrant III.

Step 1: Determine missing trigonometric values for $s$ and $t$.

For $s$:

Using the Pythagorean identity: $\sin^2(s) + \cos^2(s) = 1$ Substitute $\cos(s) = -\frac{3}{5}$: $\sin^2(s) + \left(-\frac{3}{5}\right)^2 = 1$ $\sin^2(s) + \frac{9}{25} = 1$ $\sin^2(s) = 1 - \frac{9}{25} = \frac{16}{25}$ \sin(s) = -\frac{4}{5} \quad \text{(since s is in Quadrant III, where sine is negative)}.

For $t$:

Using the Pythagorean identity: $\sin^2(t) + \cos^2(t) = 1$ Substitute $\sin(t) = -\frac{12}{13}$: $\left(-\frac{12}{13}\right)^2 + \cos^2(t) = 1$ $\frac{144}{169} + \cos^2(t) = 1$ $\cos^2(t) = 1 - \frac{144}{169} = \frac{25}{169}$ \cos(t) = -\frac{5}{13} \quad \text{(since t is in Quadrant III, where cosine is negative)}.

Step 2: Use the sum formulas to find the requested values.

(a) $\sin(s + t)$:

$\sin(s + t) = \sin(s)\cos(t) + \cos(s)\sin(t)$ Substitute the known values: $\sin(s + t) = \left(-\frac{4}{5}\right)\left(-\frac{5}{13}\right) + \left(-\frac{3}{5}\right)\left(-\frac{12}{13}\right)$ $\sin(s + t) = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}.$

(b) $\tan(s + t)$:

$\tan(s + t) = \frac{\tan(s) + \tan(t)}{1 - \tan(s)\tan(t)}.$ First, find $\tan(s)$ and $\tan(t)$: $\tan(s) = \frac{\sin(s)}{\cos(s)} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3},$ $\tan(t) = \frac{\sin(t)}{\cos(t)} = \frac{-\frac{12}{13}}{-\frac{5}{13}} = \frac{12}{5}.$ Now substitute into the formula: $\tan(s + t) = \frac{\frac{4}{3} + \frac{12}{5}}{1 - \left(\frac{4}{3} \cdot \frac{12}{5}\right)}.$ Simplify the numerator: $\frac{4}{3} + \frac{12}{5} = \frac{20}{15} + \frac{36}{15} = \frac{56}{15}.$ Simplify the denominator: $1 - \left(\frac{4}{3} \cdot \frac{12}{5}\right) = 1 - \frac{48}{15} = \frac{15}{15} - \frac{48}{15} = -\frac{33}{15}.$ Thus: $\tan(s + t) = \frac{\frac{56}{15}}{-\frac{33}{15}} = -\frac{56}{33}.$

(c) Quadrant of $s + t$:

Since both $s$ and $t$ are in Quadrant III, adding two angles in the same quadrant results in an angle in Quadrant II ($180^\circ < s + t < 270^\circ$).

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Final Answers:

1. $\sin(s + t) = \frac{56}{65}$,
2. $\tan(s + t) = -\frac{56}{33}$,
3. Quadrant of $s + t$: Quadrant II.

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Do you have questions or need further clarification?

Related Questions:

1. What is the cosine of $s + t$?
2. How would the results change if $s$ and $t$ were in Quadrant IV instead of III?
3. Can you verify $\tan(s + t)$ using alternative identities?
4. How do you determine the reference angle for $s + t$?
5. What are the exact decimal values of $\sin(s + t)$ and $\tan(s + t)$?

Tip:

Always confirm the signs of trigonometric values based on the quadrant to avoid calculation errors.