Find $\alpha$ from $\cot \alpha = 1$ and its quadrant

Given: $\cot \alpha = 1 \implies \tan \alpha = 1$

The reference angle for $\tan \alpha = 1$ is $\frac{\pi}{4}$.

Also, $\pi < \alpha < \frac{3\pi}{2}$ so $\alpha$ lies in the IIIrd quadrant.

Hence, $\alpha = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$

Therefore, $\tan \alpha = 1$

Find trigonometric ratios for $\beta$

Given: $\sec \beta = -\frac{5}{3}$ so $\cos \beta = -\frac{3}{5}$

Also, $\frac{\pi}{2} < \beta < \pi$ so $\beta$ lies in the IInd quadrant, where sine is positive.

Thus, $\sin \beta = \sqrt{1-\cos^2\beta} = \sqrt{1-\frac{9}{25}} = \frac{4}{5}$

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

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

Using the identity: $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

Substitute values: $\tan(\alpha+\beta)=\frac{1+\left(-\frac{4}{3}\right)}{1-(1)\left(-\frac{4}{3}\right)}$

$=\frac{1-\frac{4}{3}}{1+\frac{4}{3}}$

$=\frac{-\frac{1}{3}}{\frac{7}{3}} = -\frac{1}{7}$

Determine the quadrant of $\alpha+\beta$

We have: $\alpha = \frac{5\pi}{4}, \qquad \beta \in \left(\frac{\pi}{2},\pi\right)$

So,

Match with the options

The value and quadrant are: $\tan(\alpha+\beta)=-\frac{1}{7}, \quad \alpha+\beta \text{ lies in IVth quadrant}$

So the correct option is: A