If $$\cot \alpha = 1$$ and $$\sec \beta = -\frac{5}{3}$$, where $$\pi < \alpha < \frac{3\pi}{2}$$ and $$\frac{\pi}{2} < \beta < \pi$$, then the value of $$\tan(\alpha + \beta)$$ and the quadrant in which $$\alpha + \beta$$ lies, respectively are

Given: $$\cot \alpha = 1$$ with $$\pi < \alpha < \frac{3\pi}{2}$$, and $$\sec \beta = -\frac{5}{3}$$ with $$\frac{\pi}{2} < \beta < \pi$$.

Find $$\tan \alpha$$ and $$\tan \beta$$:

Since $$\cot \alpha = 1$$, we get $$\tan \alpha = 1$$.

(In the third quadrant, both sine and cosine are negative, so tangent is positive. This is consistent.)

Since $$\sec \beta = -\frac{5}{3}$$, we get $$\cos \beta = -\frac{3}{5}$$.

In the second quadrant, $$\sin \beta > 0$$, so $$\sin \beta = \frac{4}{5}$$.

$$\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/5}{-3/5} = -\frac{4}{3}$$

Compute $$\tan(\alpha + \beta)$$:

$$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta} = \frac{1 + \left(-\frac{4}{3}\right)}{1 - (1)\left(-\frac{4}{3}\right)} = \frac{-\frac{1}{3}}{1 + \frac{4}{3}} = \frac{-\frac{1}{3}}{\frac{7}{3}} = -\frac{1}{7}$$

Determine the quadrant of $$(\alpha + \beta)$$:

Since $$\alpha = \frac{5\pi}{4}$$ (the angle in the third quadrant with $$\tan \alpha = 1$$) and $$\beta = \pi - \arctan\!\left(\frac{4}{3}\right)$$ (in the second quadrant), we get:

$$\alpha + \beta = \frac{5\pi}{4} + \pi - \arctan\!\left(\frac{4}{3}\right) \approx 3.927 + 2.214 \approx 6.14$$

Since $$\frac{3\pi}{2} \approx 4.71$$ and $$2\pi \approx 6.28$$, we have $$\frac{3\pi}{2} < \alpha + \beta < 2\pi$$.

This places $$\alpha + \beta$$ in the fourth quadrant, which is consistent with $$\tan(\alpha + \beta) = -\frac{1}{7} < 0$$.

The correct answer is Option A: $$-\dfrac{1}{7}$$ and IV$$^{\text{th}}$$ quadrant.