If `cot alpha=(1)/(2) , sec beta=(-5)/(3)` where `pi lt alpha lt (3pi)/2` and `(pi)/(2) lt beta lt pi` Find the value of `tan(alpha +beta)`

Answer

To find the value of \( \tan(\alpha + \beta) \) given \( \cot \alpha = \frac{1}{2} \) and \( \sec \beta = -\frac{5}{3} \) with the specified ranges for \( \alpha \) and \( \beta \), we will follow these steps:

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

### Step 2: Find \( \tan \beta \)
Given \( \sec \beta = -\frac{5}{3} \), we can find \( \cos \beta \):
\[ \cos \beta = \frac{1}{\sec \beta} = -\frac{3}{5} \]
Now, we use the Pythagorean identity to find \( \sin \beta \):
\[ \sin^2 \beta + \cos^2 \beta = 1 \]
\[ \sin^2 \beta + \left(-\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2 \beta + \frac{9}{25} = 1 \]
\[ \sin^2 \beta = 1 - \frac{9}{25} = \frac{16}{25} \]
Thus, \( \sin \beta = \pm \frac{4}{5} \). Since \( \beta \) is in the second quadrant, \( \sin \beta \) 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 3: 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:
\[ \tan(\alpha + \beta) = \frac{2 + \left(-\frac{4}{3}\right)}{1 - (2) \left(-\frac{4}{3}\right)} \]
Calculating the numerator:
\[ 2 - \frac{4}{3} = \frac{6}{3} - \frac{4}{3} = \frac{2}{3} \]
Calculating the denominator:
\[ 1 + \frac{8}{3} = \frac{3}{3} + \frac{8}{3} = \frac{11}{3} \]
Now, substituting back into the formula:
\[ \tan(\alpha + \beta) = \frac{\frac{2}{3}}{\frac{11}{3}} = \frac{2}{11} \]

### Final Answer
Thus, the value of \( \tan(\alpha + \beta) \) is:
\[ \frac{2}{11} \]