If `cot alpha=1/2, alpha in (pi,(3pi)/2)` and `sec beta=(-5)/3, beta in ((pi)/2, pi)` find `tan (alpha+beta)`

Answer

To solve the problem, we need to find \( \tan(\alpha + \beta) \) given that \( \cot \alpha = \frac{1}{2} \) and \( \sec \beta = -\frac{5}{3} \).

### Step 1: Find \( \tan \alpha \)
Given \( \cot \alpha = \frac{1}{2} \), we know that:
\[ \tan \alpha = \frac{1}{\cot \alpha} = \frac{1}{\frac{1}{2}} = 2 \]
Since \( \alpha \) is in the range \( (\pi, \frac{3\pi}{2}) \), \( \tan \alpha \) is positive.

**Hint:** Remember that \( \tan \) is the reciprocal of \( \cot \).

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

**Hint:** Use the Pythagorean identity to find \( \sin \beta \) after determining \( \cos \beta \).

### Step 3: Calculate \( \tan(\alpha + \beta) \)
Using the formula for \( \tan(\alpha + \beta) \):
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
Substituting the values:
\[ \tan(\alpha + \beta) = \frac{2 + \frac{4}{3}}{1 - 2 \cdot \frac{4}{3}} \]
Calculating the numerator:
\[ 2 + \frac{4}{3} = \frac{6}{3} + \frac{4}{3} = \frac{10}{3} \]
Calculating the denominator:
\[ 1 - 2 \cdot \frac{4}{3} = 1 - \frac{8}{3} = \frac{3}{3} - \frac{8}{3} = -\frac{5}{3} \]
Now substituting back:
\[ \tan(\alpha + \beta) = \frac{\frac{10}{3}}{-\frac{5}{3}} = -2 \]

### Final Result
Thus, the value of \( \tan(\alpha + \beta) \) is:
\[ \boxed{-2} \]