Use the given information to find ​(a) sine left parenthesis s plus t right parenthesis​, ​(b) tangent left parenthesis s plus t right parenthesis​, and ​(c) the quadrant of s plus t.
cosine s equals negative three fifths
and sine t equals negative StartFraction 12 Over 13 EndFraction
​, s and t in quadrant III

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** (\(180^\circ < s + t < 270^\circ\)).

---

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

---

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.

Use the given information to find (a) sine(s + t), (b) tangent(s + t), and (c) the quadrant of s + t. Cosine(s) = -3/5 and Sine(t) = -12/13, with s and t in Quadrant III.

Learn how to solve for sine(s + t), tangent(s + t), and determine the quadrant of s + t using trigonometric identities. This problem involves Pythagorean identities, sum formulas, and quadrant analysis, suitable for Grades 11-12.

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