Find the value of `"tan"(alpha+beta)`given that: `cotalpha=(1)/(2) ,alphai(pi,(3pi)/(2))` and `secbeta=-(5)/(3),betain((pi)/(2),pi)` and state the quadrant in which `(alpha+beta)` terminates

Answer

To find the value of \( \tan(\alpha + \beta) \) given \( \cot \alpha = \frac{1}{2} \) where \( \alpha \) is in the interval \( \left(\frac{\pi}{3}, \frac{3\pi}{2}\right) \) and \( \sec \beta = -\frac{5}{3} \) where \( \beta \) is in the interval \( \left(\frac{\pi}{2}, \pi\right) \), we can follow these steps:

### Step 1: Find \( \tan \alpha \)
Given \( \cot \alpha = \frac{1}{2} \), we know that:
\[ \tan \alpha = \frac{1}{\cot \alpha} = \frac{1}{\frac{1}{2}} = 2 \]

### Step 2: Determine the quadrant for \( \alpha \)
Since \( \alpha \) is in the interval \( \left(\frac{\pi}{3}, \frac{3\pi}{2}\right) \), it is in the third quadrant where \( \tan \) is positive.

### Step 3: Find \( \tan \beta \)
Given \( \sec \beta = -\frac{5}{3} \), we can find \( \cos \beta \):
\[ \cos \beta = \frac{1}{\sec \beta} = -\frac{3}{5} \]
Now, using the Pythagorean identity \( \sin^2 \beta + \cos^2 \beta = 1 \):
\[ \sin^2 \beta = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \]
Thus, \( \sin \beta = \pm \frac{4}{5} \). Since \( \beta \) is in the interval \( \left(\frac{\pi}{2}, \pi\right) \), \( \sin \beta \) is positive:
\[ \sin \beta = \frac{4}{5} \]
Now, we can find \( \tan \beta \):
\[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3} \]

### Step 4: Use the formula for \( \tan(\alpha + \beta) \)
The formula for \( \tan(\alpha + \beta) \) is:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
Substituting the values we found:
\[ \tan(\alpha + \beta) = \frac{2 - \frac{4}{3}}{1 - (2)(-\frac{4}{3})} \]

### Step 5: Simplify the expression
Calculating the numerator:
\[ 2 - \frac{4}{3} = \frac{6}{3} - \frac{4}{3} = \frac{2}{3} \]
Calculating the denominator:
\[ 1 + \frac{8}{3} = \frac{3}{3} + \frac{8}{3} = \frac{11}{3} \]
Thus,
\[ \tan(\alpha + \beta) = \frac{\frac{2}{3}}{\frac{11}{3}} = \frac{2}{11} \]

### Step 6: Determine the quadrant for \( \alpha + \beta \)
Since \( \alpha \) is in the third quadrant and \( \beta \) is in the second quadrant, \( \alpha + \beta \) will be in the fourth quadrant (as \( \frac{3\pi}{2} + \frac{\pi}{2} = 2\pi \)).

### Final Answer
The value of \( \tan(\alpha + \beta) \) is \( \frac{2}{11} \) and \( \alpha + \beta \) terminates in the fourth quadrant.