If `cot alpha= 1 and sec beta = - (5)/(3)`, where `pi lt alpha lt (3pi)/(2) and (pi)/(2) lt beta lt pi`, then the value of `tan (alpha + beta)` and the quadrant in which `alpha + beta` lies, respectively are:

Options

A. `- (1)/(7) and IV^(th)` quadrant
B. `7 and I^(st)` quadrant
C. `-7 and IV^(th)` quadrant
D. `(1)/(7) and I^(st)` quadrant

Answer

To solve the problem, we need to find the value of \( \tan(\alpha + \beta) \) given that \( \cot \alpha = 1 \) and \( \sec \beta = -\frac{5}{3} \), along with the specified ranges for \( \alpha \) and \( \beta \).

### Step 1: Determine the angle \( \alpha \)

Given \( \cot \alpha = 1 \), we know that:
\[ \tan \alpha = \frac{1}{\cot \alpha} = 1 \]
The angle \( \alpha \) where \( \tan \alpha = 1 \) is \( \frac{\pi}{4} + n\pi \). Since \( \pi < \alpha < \frac{3\pi}{2} \), we take:
\[ \alpha = \frac{5\pi}{4} \quad \text{(which is in the third quadrant)} \]

### Step 2: Determine the angle \( \beta \)

Given \( \sec \beta = -\frac{5}{3} \), we can find \( \cos \beta \):
\[ \cos \beta = -\frac{3}{5} \]
Since \( \beta \) is in the second quadrant, we can find \( \sin \beta \) using the Pythagorean identity:
\[ \sin^2 \beta + \cos^2 \beta = 1 \]
Substituting for \( \cos \beta \):
\[ \sin^2 \beta + \left(-\frac{3}{5}\right)^2 = 1 \\ \sin^2 \beta + \frac{9}{25} = 1 \\ \sin^2 \beta = 1 - \frac{9}{25} = \frac{16}{25} \\ \sin \beta = \frac{4}{5} \quad \text{(positive in the second quadrant)} \]

### Step 3: Calculate \( \tan \alpha \) and \( \tan \beta \)

Now we can find \( \tan \alpha \) and \( \tan \beta \):
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{1}{1} = 1 \quad \text{(from Step 1)} \]
For \( \tan \beta \):
\[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3} \]

### Step 4: Use the tangent addition formula

Now we can use the formula for \( \tan(\alpha + \beta) \):
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]
Substituting the values:
\[ \tan(\alpha + \beta) = \frac{1 + \left(-\frac{4}{3}\right)}{1 - (1)(-\frac{4}{3})} \\ = \frac{1 - \frac{4}{3}}{1 + \frac{4}{3}} \\ = \frac{\frac{3}{3} - \frac{4}{3}}{\frac{3}{3} + \frac{4}{3}} \\ = \frac{-\frac{1}{3}}{\frac{7}{3}} \\ = -\frac{1}{7} \]

### Step 5: Determine the quadrant of \( \alpha + \beta \)

Since \( \alpha = \frac{5\pi}{4} \) (third quadrant) and \( \beta \) is in the second quadrant, we can estimate \( \alpha + \beta \):
\[ \alpha + \beta = \frac{5\pi}{4} + \beta \]
Since \( \beta \) is between \( \frac{\pi}{2} \) and \( \pi \), we can estimate:
\[ \frac{5\pi}{4} + \frac{\pi}{2} < \alpha + \beta < \frac{5\pi}{4} + \pi \\ \frac{5\pi + 2\pi}{4} < \alpha + \beta < \frac{5\pi + 4\pi}{4} \\ \frac{7\pi}{4} < \alpha + \beta < \frac{9\pi}{4} \]
This implies that \( \alpha + \beta \) is in the fourth quadrant.

### Final Answer

Thus, the value of \( \tan(\alpha + \beta) \) is \( -\frac{1}{7} \) and \( \alpha + \beta \) lies in the fourth quadrant.