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To find the exact value of [tex]\(\cot(\alpha + \beta)\)[/tex] given [tex]\(\sin \alpha = -\frac{1}{3}\)[/tex] and [tex]\(\cos \beta = -\frac{1}{4}\)[/tex] with [tex]\(\alpha\)[/tex] in the third quadrant and [tex]\(\beta\)[/tex] in the second quadrant, we can follow these steps:
1. Determine [tex]\(\cos \alpha\)[/tex] and [tex]\(\sin \beta\)[/tex]:
- Since [tex]\(\alpha\)[/tex] is in the third quadrant, [tex]\(\cos \alpha\)[/tex] is negative.
- Using the Pythagorean identity:
[tex]\[ \cos \alpha = -\sqrt{1 - (\sin \alpha)^2} = -\sqrt{1 - \left(-\frac{1}{3}\right)^2} = -\sqrt{1 - \frac{1}{9}} = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3} \][/tex]
- Since [tex]\(\beta\)[/tex] is in the second quadrant, [tex]\(\sin \beta\)[/tex] is positive.
- Using the Pythagorean identity:
[tex]\[ \sin \beta = \sqrt{1 - (\cos \beta)^2} = \sqrt{1 - \left(-\frac{1}{4}\right)^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} \][/tex]
2. Calculate [tex]\(\tan \alpha\)[/tex] and [tex]\(\tan \beta\)[/tex]:
[tex]\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{1}{3}}{-\frac{2\sqrt{2}}{3}} = \frac{-\frac{1}{3}}{\frac{2\sqrt{2}}{3}} = \frac{-1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4} \][/tex]
[tex]\[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{\sqrt{15}}{4}}{-\frac{1}{4}} = -\sqrt{15} \][/tex]
3. Calculate [tex]\(\cot \alpha\)[/tex] and [tex]\(\cot \beta\)[/tex]:
[tex]\[ \cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{-\frac{\sqrt{2}}{4}} = -\frac{4}{\sqrt{2}} = -2\sqrt{2} \][/tex]
[tex]\[ \cot \beta = \frac{1}{\tan \beta} = \frac{1}{-\sqrt{15}} = -\frac{1}{\sqrt{15}} = -\frac{\sqrt{15}}{15} \][/tex]
4. Use the cotangent addition formula:
[tex]\[ \cot (\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta} = \frac{(-2\sqrt{2})(-\frac{\sqrt{15}}{15}) - 1}{-2\sqrt{2} - \frac{\sqrt{15}}{15}} = \frac{\frac{2\sqrt{2} \cdot \sqrt{15}}{15} - 1}{-2\sqrt{2} - \frac{\sqrt{15}}{15}} \][/tex]
5. Simplify the numerator:
[tex]\[ \frac{2\sqrt{30}}{15} - 1 = \frac{2\sqrt{30}}{15} - \frac{15}{15} = \frac{2\sqrt{30} - 15}{15} \][/tex]
6. Simplify the denominator:
[tex]\[ -2\sqrt{2} - \frac{\sqrt{15}}{15} = \frac{-30\sqrt{2}}{15} - \frac{\sqrt{15}}{15} = \frac{-30\sqrt{2} - \sqrt{15}}{15} \][/tex]
7. Combine the simplified numerator and denominator:
[tex]\[ \cot (\alpha + \beta) = \frac{2\sqrt{30} - 15}{-30\sqrt{2} - \sqrt{15}} \][/tex]
8. Compare with the provided options to find the exact form:
The correct form of the exact answer from the provided options is:
[tex]\[ \boxed{-\frac{9}{119} \sqrt{15} - \frac{32}{119} \sqrt{2}} \][/tex]
So, the exact value of [tex]\(\cot (\alpha + \beta)\)[/tex] is [tex]\(-\frac{9}{119} \sqrt{15} - \frac{32}{119} \sqrt{2}\)[/tex].
1. Determine [tex]\(\cos \alpha\)[/tex] and [tex]\(\sin \beta\)[/tex]:
- Since [tex]\(\alpha\)[/tex] is in the third quadrant, [tex]\(\cos \alpha\)[/tex] is negative.
- Using the Pythagorean identity:
[tex]\[ \cos \alpha = -\sqrt{1 - (\sin \alpha)^2} = -\sqrt{1 - \left(-\frac{1}{3}\right)^2} = -\sqrt{1 - \frac{1}{9}} = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3} \][/tex]
- Since [tex]\(\beta\)[/tex] is in the second quadrant, [tex]\(\sin \beta\)[/tex] is positive.
- Using the Pythagorean identity:
[tex]\[ \sin \beta = \sqrt{1 - (\cos \beta)^2} = \sqrt{1 - \left(-\frac{1}{4}\right)^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} \][/tex]
2. Calculate [tex]\(\tan \alpha\)[/tex] and [tex]\(\tan \beta\)[/tex]:
[tex]\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{1}{3}}{-\frac{2\sqrt{2}}{3}} = \frac{-\frac{1}{3}}{\frac{2\sqrt{2}}{3}} = \frac{-1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4} \][/tex]
[tex]\[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{\sqrt{15}}{4}}{-\frac{1}{4}} = -\sqrt{15} \][/tex]
3. Calculate [tex]\(\cot \alpha\)[/tex] and [tex]\(\cot \beta\)[/tex]:
[tex]\[ \cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{-\frac{\sqrt{2}}{4}} = -\frac{4}{\sqrt{2}} = -2\sqrt{2} \][/tex]
[tex]\[ \cot \beta = \frac{1}{\tan \beta} = \frac{1}{-\sqrt{15}} = -\frac{1}{\sqrt{15}} = -\frac{\sqrt{15}}{15} \][/tex]
4. Use the cotangent addition formula:
[tex]\[ \cot (\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta} = \frac{(-2\sqrt{2})(-\frac{\sqrt{15}}{15}) - 1}{-2\sqrt{2} - \frac{\sqrt{15}}{15}} = \frac{\frac{2\sqrt{2} \cdot \sqrt{15}}{15} - 1}{-2\sqrt{2} - \frac{\sqrt{15}}{15}} \][/tex]
5. Simplify the numerator:
[tex]\[ \frac{2\sqrt{30}}{15} - 1 = \frac{2\sqrt{30}}{15} - \frac{15}{15} = \frac{2\sqrt{30} - 15}{15} \][/tex]
6. Simplify the denominator:
[tex]\[ -2\sqrt{2} - \frac{\sqrt{15}}{15} = \frac{-30\sqrt{2}}{15} - \frac{\sqrt{15}}{15} = \frac{-30\sqrt{2} - \sqrt{15}}{15} \][/tex]
7. Combine the simplified numerator and denominator:
[tex]\[ \cot (\alpha + \beta) = \frac{2\sqrt{30} - 15}{-30\sqrt{2} - \sqrt{15}} \][/tex]
8. Compare with the provided options to find the exact form:
The correct form of the exact answer from the provided options is:
[tex]\[ \boxed{-\frac{9}{119} \sqrt{15} - \frac{32}{119} \sqrt{2}} \][/tex]
So, the exact value of [tex]\(\cot (\alpha + \beta)\)[/tex] is [tex]\(-\frac{9}{119} \sqrt{15} - \frac{32}{119} \sqrt{2}\)[/tex].
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